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Polynomial LongDivision What would be the result of $x^3-4x^2-5$ divided by $x-3$ ? I am getting $4$ as my solution can someone prove me wrong, this is very confusing.
|
The remainder when dividing by $x-3$ is also the polynomial evaluated at $x=3$, which is $27-4\cdot 9-5=-14$.
Indeed,
$$\begin{align}x^3-4x^2-5&=x^2\cdot(x-3)-x^2-5\\&=(x^2-x)(x-3)-3x-5\\&=(x^2-x-3)(x-3)-14\end{align}$$
|
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|
$\sum _{n=0}^{\infty } \frac{x^n}{n+3}$, sum, area of convergence & center I want to find the
$$\sum _{n=0}^{\infty } \frac{x^n}{n+3}$$
now this is how I thought about doing it but I get stuck.
$$\sum _{n=0}^{\infty } x^n=\frac{1}{1-x}$$ given that absolute value of x is less then zero.
$$\int x^n \, dx=\frac{x^{n+1}}{n+1}$$
I keep integrating the result until I get
$$\frac{x^{n+3}}{(n+3) (n+1) (n+2)}=\frac{3 x^2}{4}-\frac{1}{2} x^2 \log (1-x)-\frac{x}{2}+x \log (1-x)-\frac{1}{2} \log (1-x)$$
I thought about multiplying by $$\frac{(n+1) (n+2)}{x^3}$$
However, the answer should be this: $$\sum _{n=0}^{\infty } \frac{x^n}{n+3}=-\frac{\log (1-x)}{x^3}-\frac{1}{x^2}-\frac{1}{2 x}$$
On the interval [-1,1), and $\frac{1}{3}$ if x=0.
Could someone please show me what I did wrong and how the steps should be to get the correct answer?
|
I think this is a good way of handling these series:
Let $s(x)=\sum_{n=0}^{+\infty} \frac{x^n}{n+3}$ (defined for $|x|<1$). Then
$$
x^3s(x)=\sum_{n=0}^{+\infty}\frac{x^{n+3}}{n+3},
$$
and hence
$$
(x^3s(x))'=\sum_{n=0}^{+\infty}x^{n+2}=x^2\sum_{n=0}^{+\infty}x^n=\frac{x^2}{1-x}.
$$
Integrating from $0$ to $t$, we get
$$
t^3s(t)-0^3s(0)=\int_0^t\frac{x^2}{1-x}\,dx = -t-\frac{t^2}{2}-\log(1-t),
$$
so (switching back to $x$ as variable)
$$
s(x)=-\frac{1}{x^2}-\frac{1}{2x}-\frac{\log(1-x)}{x^3}.
$$
|
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|
How to solve this limit: $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$ Compute the limit $\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$
my attempt:
I tried to multiply top and bottom by the conjugate
$$\begin{align}
\lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-1}\right)\frac{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{\left(\sqrt{x+\sqrt{x}}\right)^2-\left(\sqrt{x-1}\right)^2}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{(x+\sqrt{x})-(x-1)}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}
\end{align}$$
But I don't know what I can do after this.
|
Let's start from your last line:
$$\begin{align}
\lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}} &= \lim \frac{\sqrt x}{\sqrt x} \frac{\frac{1}{\sqrt x} + 1}{\sqrt{1 + \frac{1}{\sqrt x}} + \sqrt{1 - \frac{1}{ x}}} \\
&= \frac{1}{1 + 1} = \frac{1}{2}
\end{align}$$
where we note that everywhere we have $\frac{1}{\sqrt x}$, those terms go to $0$ as $x \to \infty$. The method of factoring out the largest element in the numerator and denominator very often works. $\diamondsuit$
|
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|
If $n$ is a positive integer such that $2^n+n^2$ is a prime number , then is it true that $6|n-3$ ? If $n$ is a positive integer such that $2^n+n^2$ is a prime number , then is it true that $6|n-3$ ? Trivially $n$ cannot be even , so this leaves us only with the possibilities $n \equiv1,3,5( \mod 6) $ , but I cannot reduce the cases . Please help , Thanks in advance .
|
If $n\equiv1,5\pmod 6$ then $n^2\equiv 1\pmod 6$.
On the other hand, $2^n\equiv 2$ or $4\pmod 6$ depending on $n$ is odd or even. So if $n^2\equiv 1\pmod 6$, then
$$2^n+n^2\equiv 2+1\equiv 3\pmod 6$$
therefore $2^n+n^2$ is a multiple of three.
Notice the case $1^2+2^1$, that is the only exception to your statement.
|
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|
The dimension of the SU(2) matrix group Let's take the matrix $R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
Consider its transpose $R^\dagger = \begin{pmatrix} a^* & c^* \\ b^* & d^* \end{pmatrix}$.
Then $RR^\dagger =1$ implies that $|a|^2 + |c|^2 = 1, |b|^2 + |d|^2 = 1, ab^* + cd^* = 0,$ and $a^*b + c^*d = 0$.
The last two equations are equivalent to each other, so we have three complex equations.
If $a=a_0+ia_1, b=b_0+ib_1, c=c_0+ic_1,$ and $d=d_0+id_1$, then the above three complex equations lead to $a_0^2 + a_1^2+c_0^2+c_1^2=1, b_0^2 + b_1^2+d_0^2+d_1^2=1, a_{0}b_{0} + a_{1}b_{1} + c_{0}d_{0} + c_{1}d_{1} = 0,$ and $a_{0}b_{1} - a_{1}b_{0} + c_{1}d_{0} - c_{0}d_{1} = 0$.
Therefore, the three complex equations lead to four real equations (the first two equations have only real components, but the final equation has both real and complex components).
Therefore, the four real equations ensure that the number of real variables that parametrise the matrix $R$ drops from 8 to 4.
Now, let's consider the determinant equation $ad-bc=1$, which decomposes into $(a_{0}d_{0} - a_{1}d_{1} - b_{0}c_{0} + b_{1}c_{1}) + i(a_{1}d_{0} + a_{0}d_{1} - b_{1}c_{0} - b_{0}c_{1}) = 1,$ which decomposes into $a_{0}d_{0} - a_{1}d_{1} - b_{0}c_{0} + b_{1}c_{1} = 1,$ and $a_{1}d_{0} + a_{0}d_{1} - b_{1}c_{0} - b_{0}c_{1} = 0$.
Therefore, we have two real equations, so the number of parameters that characterise the matrix $R$ should now drop from 4 to 2. But several sources mention that the number of parameters drop from 4 to 3.
Can someone explain?
|
The explanation is that the determinant is already constrained to be on the unit circle. We have $|\det(R)|^2 = \det(R)\overline{\det(R)} = \det(R)\det(R^\dagger) = \det(RR^\dagger) =\det(I) = 1$.
This means that the equation $a_0d_0 - a_1d_1 - b_0c_0 + b_1c_1 = 1$ implies $a_0d_1 + a_1d_0 - b_0c_1 - b_1c_0 = 0$, and the latter equation implies $a_0d_0 - a_1d_1 - b_0c_0 + b_1c_1 = \pm 1$. In other words, the set of five real equations has a nontrivial relation, which increases the dimension of the solution set.
Another way of looking at it is that the determinant gives a homomorphism $U(2) \to S^1$, and the first isomorphism theorem then guarantees that the dimension of the kernel $SU(2)$ is the dimension of the source minus the dimension of the image.
|
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|
Sum of square patterns Can anyone give the name of this pattern
$$136^2+137^2+138^2+139^2+140^2+141^2+142^2+143^2+144^2 =\\
145^2+146^2+147^2+148^2+149^2+150^2+151^2+152^2$$
|
It belongs to a family that starts with the familiar,
$$\begin{align}
&3^2+4^2 = 5^2\\
&10^2+11^2+12^2=13^2+14^2\\
&21^2+22^2+23^2+24^2 = 25^2+26^2+27^2\\
\vdots\\
&a^2 + (a+1)^2 + \dots + a_n^2 = b^2 + (b+1)^2 + \dots + b_{n-1}^2
\end{align}$$
where $b = a_n+1$ and $a = 2n^2-3n+1 = 0, 3, 10, 21, 36, 55, 78, 105, \color{brown}{136},\dots$ Yours was the case of $n=9$ squares. In general, since,
$$a^2+(a+1)^2+\dots+(a+p-1)^2 = (p/6)(6a^2-6a + 6ap + 1 - 3p + 2p^2)$$
then you need to solve an equation of the form,
$$p(6a^2-6a + 6ap + 1 - 3p + 2p^2) = q(6b^2 - 6b + 6b q + 1 - 3q + 2q^2)\tag1$$
in positive integers. A parametric solution to $(1)$ is,
$$a = 2p^2-3p+1$$
$$b = a+p$$
$$q = p-1$$
for any $p$. The "continuous" case $b=a+p$ like the family above also has sporadic solutions like the $a,p,q=4,35,10$ given by bobbym, though these do not seem to belong to a second family.
|
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|
Show that the substitution $t=\tan\theta$ transforms the integral ${\int}\frac{d\theta}{9\cos^2\theta+\sin^2\theta}$, into ${\int}\frac{dt}{9+t^2}$ To begin with the $d\theta$ on the top of the fraction threw me off but I'm assuming it's just another way of representing:
$${\int}\frac{1}{9\cos^2\theta+\sin^2\theta}\,d\theta$$
I tried working backwards
$$\frac{d}{d\theta}\tan\theta=\sec^2\theta\,\,\,\,{\Rightarrow}\,\,\,\,d\,\tan\theta=\sec^2\theta\,d\theta$$
$${\Rightarrow}\,{\int}\frac{\sec^2\theta\,d\theta}{9+\tan^2\theta}$$
$$\tan\theta=\frac{\sin\theta}{\cos\theta}\,\,\,\,{\Rightarrow}\,\,\,\,\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}$$
$${\Rightarrow}\,{\int}\frac{\sec^2\theta\,d\theta}{\left(9+\dfrac{\sin^2\theta}{\cos^2\theta}\right)}$$
$$9=\frac{9\cos^2\theta}{\cos^2\theta}\,\,\,\,{\Rightarrow}\,\,\,\,{\int}\frac{\sec^2\theta\,d\theta}{\left(\dfrac{9\cos^2\theta}{\cos^2\theta}+\dfrac{\sin^2\theta}{\cos^2\theta}\right)}\,\,\,\,{\Rightarrow}\,\,\,\,{\int}\frac{\sec^2\theta\,d\theta}{\left(\dfrac{9\cos^2\theta+\sin^2\theta}{\cos^2\theta}\right)}$$
$${\Rightarrow}\,\,\,\,{\int}\frac{\color{red}{\cos^2\theta\,\sec^2\theta}\,d\theta}{9\cos^2\theta+\sin^2\theta}$$
Now I have to prove that $$\cos^2\theta\,\sec^2\theta=1$$
but I don't think it is... What have I done wrong? Regards Tom
|
An alternative method. Let
$$I={\int}\frac{1}{9\cos^2x+\sin^2x}\,dx$$
Dividing the denominator and numerator by $\cos^2x$ we have
$$I=\int\frac{\sec^2x}{9+\tan^2x}dx$$
Make the substitution $t=\tan x$, so that $\sec^2x~dx=dt,$ and we are done.
|
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|
Odd divisibility induction proof Prove that for odd $n>3$
$$64\ | \ n^4-18n^2+17$$
I checked that for $n=5$ it works. I think I need to assume that for $2n+1$ it holds and show that $2n+3$ also holds. Any ideas?
|
If $n=2k+1$, we have
$$
\begin{align}
n^4-18n^2+17
&=(2k+1)^4-18(2k+1)^2+17\\[6pt]
&=16k^4+32k^3-48k^2-64k\\
&=384\binom{k}{4}+768\binom{k}{3}+320\binom{k}{2}-64\binom{k}{1}\\
&=64\left[6\binom{k}{4}+12\binom{k}{3}+5\binom{k}{2}-\binom{k}{1}\right]
\end{align}
$$
This indicates that $64$ divides $n^4-18n^2+17$ for all odd $n$.
|
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|
Comparing series Can anyone explain why if I compare the coefficient of $x^{n}$ of the equation
$$\sum_{k=0}^{\infty}a(n)x^n= \frac{1}{1-x}-\frac{x}{1-x^3}+\frac{x^2}{1-x^5}-\frac{x^3}{1-x^7}+...$$
I can get
$$a(n)=k_{1}(4n+1)-k_{3}(4n+1)$$ where $k_{i}(m)$ is the number of divisors of $m$ that are congruent to $j$ modulo $4$?
|
$$
\begin{align}
\sum_{n=0}^\infty(-1)^n\frac{x^n}{1-x^{2n+1}}
&=\sum_{n=0}^\infty\sum_{k=0}^\infty(-1)^nx^{n+(2n+1)k}\\
\end{align}
$$
The coefficient of $x^m$ is the the number of factors of $2m+1=(2n+1)(2k+1)$ that are $1$ mod $4$ ($n$ even) minus the number that are $3$ mod $4$ ($n$ odd). This confirms the formula in the question.
Each prime factor, $p$, of $2m+1$ that is $3$ mod $4$, which has an exponent of $k$, contributes to this difference of factor counts, a factor of
$$
\overbrace{1-1+1-1+\cdots+(-1)^k}^{\text{$k+1$ terms}}
$$
which is $1$ if $k$ is even and $0$ if $k$ is odd.
Each prime factor, $p$, of $2m+1$ that is $1$ mod $4$, which has an exponent of $k$, contributes to this difference of factor counts, a factor of
$$
\overbrace{1+1+1+1+\cdots+1}^{\text{$k+1$ terms}}
$$
which is $k+1$.
Thus, if we break $2m+1$ into $p_1$, the product of primes that are $1$ mod $4$, and $p_3$, the product of primes that are $3$ mod $4$, then the coefficient of $x^m$ is the number of factors of $p_1$ if $p_3$ is a square, and $0$ if $p_3$ is not a square.
Note that if $m$ is odd, then $2m+1\equiv3\pmod4$. This means that $2m+1$ has at least one prime factor that is $3$ mod $4$ to an odd power. Thus, the coefficient of $x^m$ is $0$ if $m$ is odd.
Here are the terms up to $x^{50}$:
$$
1 + 2 x^2 + x^4 + 2 x^6 + 2 x^8 + 3 x^{12} + 2 x^{14} + 2 x^{18} + 2 x^{20} + 2 x^{22} + x^{24}\\ + 2 x^{26} + 2 x^{30} + 4 x^{32} + 2 x^{36} + x^{40} + 4 x^{42} + 2 x^{44} + 2 x^{48} + 2 x^{50}
$$
|
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When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble - trouble understanding proof Theorem: When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble
Proof: When $p=2$, the statement is clear.
Assume $p\equiv 1\pmod{4}$, let $r=\frac{p-1}{2}$ and $x=r!$
Then since $r$ is even $x^2\equiv (1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\equiv (p-1)!\equiv -1\pmod{p}$
Thus, when $p\equiv 1\pmod{4}$, the congruence $x^2\equiv -1\pmod{p}$ is soluble.
Point of contention: I understand the general argument
I understand the relation
$(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\equiv (p-1)!\equiv -1\pmod{p}$
But I cant work out how $x^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$
How is $((\frac{p-1}{2})!)^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$
|
Since there are an even number of terms:
$$\begin{align}1\cdot 2\cdots r &= (-1)(-2)\cdots (-r)\\
&\equiv (p-1)(p-2)\cdots(p-r)\pmod p\\
&=(r+1)(r+2)\cdots(p-1)
\end{align}$$
|
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|
Solve the limit $\lim\limits _{x\to 0}\frac{\sqrt{1-\cos\left(x^2\right)}}{1-\cos\left(x\right)}$ $$\lim _{x\to \:0}\frac{\sqrt{1-\cos\left(x^2\right)}}{1-\cos\left(x\right)}=\left|\frac{0}{0}\right|$$
I think you have to multiply by the conjugate. And then make the change equivalent small. Right?
|
Remove the square root by observing that $1-\cos x\ge0$, so $1-\cos x=\sqrt{(1-\cos x)^2}$; hence you reduce to computing
$$
\lim_{x\to0}\frac{1-\cos(x^2)}{(1-\cos x)^2}
$$
and then take the square root of the result. Now
$$
\lim_{x\to0}\frac{1-\cos(x^2)}{(1-\cos x)^2}=
\lim_{x\to0}\frac{1-(1-x^4/2+o(x^4))}{(1-(1-x^2/2+o(x^2))^2}=
\lim_{x\to0}\frac{x^4/2+o(x^4)}{x^4/4+o(x^4)}=2
$$
You can even avoid using Taylor's expansion by recalling that
$$
\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}
$$
so you can directly write
$$
\lim_{x\to0}\frac{1-\cos(x^2)}{(1-\cos x)^2}=
\lim_{x\to0}\frac{1-\cos(x^2)}{x^4}\left(\frac{x^2}{1-\cos x}\right)^2=
\frac{1}{2}\cdot 4=2
$$
|
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|
System of equations with radicals: $2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}$ and $2\sqrt[4]{\frac{y^4}{3}+4} = 1+\sqrt{\frac{3}{2}x^2}$
Solve the system of equations (in $\mathbb R$):
$$\begin{matrix}
2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}
\\
2\sqrt[4]{\frac{y^4}{3}+4}
=
1+\sqrt{\frac{3}{2}x^2}
\end{matrix}.$$
This is from an older question, which was closed and deleted because of lack of context. (Here is link for users who can see deleted questions.) I will post my solution below - so I hope this time the question will not be closed for the lack of effort.
I found the system not too easy and somewhat interesting. (Of course I might have missed some straightforward way to the solution.) I'd be glad to see some other methods to solve it.
|
As you say, you will have solution corresponding to $x=\pm y$. Now, if you expand the last equation, you have (if $x \geq 0$),$$-\frac{37 x^4}{12}+3 \sqrt{6} x^3+9 x^2+2 \sqrt{6} x-63=0$$ To get rid of the $\sqrt{6}$'s, define $x=\sqrt{6} z$ and the equation becomes $$-111 z^4+108 z^3+54 z^2+12 z-63=0$$ By inspection $z=1$ is a solution. Making the long division let you with $$-111 z^3-3 z^2+51 z+63=0$$ where $z=1$ is solution again. Another division and arrive to $$-111 z^2-114 z-63=0$$ which does not show real solutions.
So, the real solutions are all possible combinations of $(\pm \sqrt{6},\pm \sqrt{6})$
|
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|
Find the coefficient of $x^{30}$. Find the coefficient of $x^{30}$ in the given polynomial
$$
\left(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}\right)^5
$$
I don't know how to solve problems with such high degree.
|
The coefficient attached to $x^{30}$ will be the number of ways you can add up to $30$ by using the numbers $0$-$12$ up to five times. (Here order matters)
For instance $1+1+2+10+6=30$ is one way.
$10+10+10+0+0=30$ is another and so is $0+10+10+10+0 = 30$.
The reason for this is more apparent for smaller polynomials.
For instance let's calculate the coefficient of $x^4$ in $(1+x+x^2)^3$.
$$(1+x+x^2)(1+x+x^2)(1+x+x^2)$$
Eventually we will have every combination of products between the three terms. We can achieve $x^4$ is several ways:
$$1\cdot x^2 \cdot x^2$$
$$x^2 \cdot 1 \cdot x^2$$
$$x^2 \cdot x^2 \cdot 1$$
$$x \cdot x \cdot x^2$$
$$x \cdot x^2 \cdot x$$
$$x^2 \cdot x \cdot x$$
Thus we have six ways of achieving $x^4$, and the powers of $x$ in each of these cases add up to $4$. Therefore the coefficient of $x^4$ is six.
|
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Proving $\left(a+\frac{2}{a}\right)^2+\left(b+\frac{2}{b}\right)^2\ge \frac{81}{2}$ for all positive real $a,b$ such that $a+b=1$ I approached this problem in two different ways, but only one was successful. I'll post the latter as an answer, while here follows the first approach:
I expanded the squares:
$$a^2+\frac{4}{a^2}+4+b^2+\frac{4}{b^2}+4\ge\frac{81}{2} \\ a^2+b^2+\frac{4}{a^2}+\frac{4}{b^2}\ge \frac{65}{2}$$ and then multiplied both sides by $a^2b^2$ to get $$a^2b^2(a^2+b^2)+4(a^2+b^2)\ge\frac{65}{2}a^2b^2 \\ (a^2+b^2)(a^2b^2+4)\ge\frac{65}{2}a^2b^2, $$ which, combining $a+b=1$ and $(a+b)^2=a^2+b^2+2ab$, can be rewritten as $$(1-2ab)(a^2b^2+4)\ge\frac{65}{2}a^2b^2.\tag{$\star$}$$ Setting $c=ab$ makes $(\star)$ an inequality in one variable, but it didn't help me. Is this totally the wrong track? And, I was wondering if there are other approaches besides what I came up with.
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The function $f: x \mapsto (x+\frac2x)^2$ is convex on $(0,\infty)$, so by Jensen's inequality we have
$$
\frac{\left(a+\frac2a\right)^2+\left(b+\frac2b\right)^2}{2} = \frac{f(a)+f(b)}{2} \geq f\left( \frac{a+b}{2} \right) = f\left(\frac12\right) = \left( \frac92 \right)^2 = \frac{81}{4}.
$$
|
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|
Complex Numbers Question, IIT JEE [2006]. Please tell me whether I solved it properly? $Q.$The value of $\sum\limits_{k=1}^{10}(\sin{\frac{2k\pi}{11}-i\cos\frac{2k\pi}{11}})$ is-?
I solved it like this-
$\frac{\sum\limits_{k=1}^{10}(\cos{\frac{2k\pi}{11}+i\sin\frac{2k\pi}{11}})}{i}$
If we observe these are the roots of the equation $z^{11}=1$
So $1+z_1+z_2+...+z_{10}=1$ (De Moivre
s Theorem)
So $z_1+z_2+...+z_{10}=-1$
$-1=i^2$
So $\frac{i^2}{i}=i$
|
Let $$\zeta_{11} = e^{2\pi i/11} = \cos \frac{2\pi}{11} + i \sin \frac{2\pi}{11}$$ be a primitive $11^{\rm th}$ root of unity; hence $$\zeta_{11}^0, \zeta_{11}^1, \ldots, \zeta_{11}^{10}$$ are the roots of $z^{11} - 1 = 0$, and the sum of these roots is therefore zero. Then
$$\begin{align*} \sum_{k=1}^{10} \left( \sin \frac{2\pi k}{11} - i \cos \frac{2\pi k}{11} \right) &= \frac{1}{i} \sum_{k=1}^{10} e^{2\pi i k/11} \\ &= \frac{1}{i}\left( -1 + \sum_{k=0}^{10} \zeta_{11}^k \right) \\ &= \frac{1}{i} (-1) \\ &= i. \end{align*}$$
To see that the sum of the aforementioned roots is zero, we could also have explicitly summed the geometric series: $$\sum_{k=0}^{n-1} \zeta_n^k = \frac{\zeta_n^n - 1}{\zeta_n - 1} = \frac{0}{\zeta_n - 1} = 0,$$ for $n > 2$.
|
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|
What is the tip for this exact differential equation? $$ xdx + ydy = \frac{xdy - ydx}{x^2 + y^2} $$
I have multiplied the left part $x^2+y^2$ for $x dx + y dy$ getting
$$(x^3+xy^2+y)dx+(x^2y+y^3-x)dy=0$$
And the derivative test give me:
$\frac{dM}{dy}= 0+2xy+1$ and $\frac{dN}{dx} = 2xy+0-1$.
Where´s my mistake?
|
When you multiplied by $x^2 + y^2$, you transformed an exact differential equation to an inexact one.
If we rearrange the original equation, we get
$$\tag{*}\left(x + \frac{y}{x^2 + y^2}\right)\, dx + \left(y - \frac{x}{x^2 + y^2}\right)\, dy = 0$$
Now
$$\frac{\partial}{\partial y}\left(x + \frac{y}{x^2 + y^2}\right) = 0 +\frac{x^2 - y^2}{(x^2 + y^2)^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2}$$
and
$$\frac{\partial}{\partial x}\left(y - \frac{x}{x^2 + y^2}\right) = 0-\frac{y^2 - x^2}{(x^2 + y^2)^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2},$$
so $(*)$ is exact and there is an $F$ such that $F_x = x + \frac{y}{x^2 + y^2}$ and $F_y = y - \frac{x}{x^2 + y^2}$. Integrating the equation $F_x = x + \frac{y}{x^2 + y^2}$ with respect to $x$, we get $F(x,y) = \frac{x^2}{2} + \arctan(x/y) + \phi(y)$ for some function $\phi(y)$ depending only on $y$. Differentiating the latter equation with respect to $y$ yields
$$F_y = -\frac{x}{x^2 + y^2} + \phi'(y) = y - \frac{x}{x^2 + y^2}.$$
Thus $\phi'(y) = y$ $\implies$ $\phi(y) = \frac{y^2}{2} + c$, where $c$ is a constant. Hence $F(x,y) = \frac{x^2+y^2}{2} + \arctan(x/y) + c$, and the general solution is
$$\frac{x^2 + y^2}{2} + \arctan(x/y) = C.$$
|
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Solution verification: Prove by induction that $a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n} $ is increasing and bounded by $2$ I have the following recursive relation (sequence):
\begin{align}
a_1 = \sqrt{2}, \quad a_{n+1} = \sqrt{2 + a_n}
\end{align}
My Try:
I'm a little skeptical of my manipulations near the end but it looks like it works out.
Base Case:
Let $n=1$ then
\begin{align}
&a_2 = \sqrt{2 + a_1} \\
&a_2 < 2 \\
&\sqrt{2 + \sqrt{2}} < 2
\end{align}
The base case holds.
Induction hypothesis:
Let $n=k$
$$a_1 = \sqrt{2} \quad a_{k+1} = \sqrt{2 + a_k} \quad a_{k+1} = \sqrt{2}$$
Induction Step:
Now we have to prove that $a_{k+2} < 2$. Let $n=k+1$.
\begin{align}
a_{k+2} &= \sqrt{2 + a_{k+1}} \\
\implies a_{k+2} &= \sqrt{2 + \sqrt{2 + a_k}} \\
\end{align}
Now we have to show that $a_{k+2} < 2$.
\begin{align}
a_{k+2} &< 2\\
\sqrt{2 + \sqrt{2 + a_k}} &< 2\\
2 + \sqrt{2 + a_k} &< 4 \\
\sqrt{2 + a_k} &< 2 \\
\end{align}
Q.E.D
Are my steps correct?
Thanks for your time!
|
I think you're making a bit of confusion. On one hand you are never proving that the sequence is increasing, on the other hand your argument is a bit too complicated. Here is how I would go about proving it:
*
*$\mathbf{n = 1}$: First note that $a_1 < a_2$ if and only if
$$
a_1^2 = 2 < 2 + \sqrt{2} = a_2^2
$$
which is clearly true. Further, $a_2 < 2$ because $2 + \sqrt{2} < 4$.
*Assume that $a_{n-1} < a_n < 2$. Then we need to prove that $a_n < a_{n+1} < 2$. For the first part note that $a_n < a_{n+1}$ is equivalent to $a_n^2 < 2 + a_n$, i.e.
$$
a_n (a_n - 1) < 2
$$
which is true because $a_n < 2$ implies $a_n - 1 < 1$. For the second part, note that $a_{n+1} < 2$ is equivalent to
$$
2 + a_n < 4
$$
which is true because $a_n < 2$ by hypothesis.
|
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|
Floor inequality with prime If $a$ and $b$ are positive integers and $a\ge b$ and $b$ is an odd prime, show that:
$$\left\lfloor \frac{6a-1}{b}\right\rfloor+\left\lfloor\frac{a}{b}\right\rfloor\ge \left\lfloor \frac{2a}{b}\right\rfloor+\left\lfloor \frac{3a-1}{b}\right\rfloor+\left\lfloor \frac{2a+1}{b}\right\rfloor$$
I tested some numbers and it looks to be true but I haven't been able to prove it.
Please help, thanks!
|
Let $a=nb+k$, with $0\le k < b$. Your inequality simplifies to
$$\left\lfloor\frac{6k-1}{b}\right\rfloor\ge\left\lfloor\frac{2k}{b}\right\rfloor+\left\lfloor\frac{3k-1}{b}\right\rfloor+\left\lfloor\frac{2k+1}{b}\right\rfloor$$
Now let's examine it case by case.
Case 1. $k = 0:$ Both sides of the inequality are $-1$, so it holds.
Case 2. $0< k < \dfrac{b+1}3:$ The left side is at least $0$ and the right side is $0$.
Case 3. $\dfrac{b+1}3\le k <\dfrac b2:$ The left side is at least $2$ and the right side is $1$.
Case 4. $\dfrac b2\le k<\dfrac{2b+1}3:$ The left side is at least $3$ and the right side is at most $3$.
Case 5. $\dfrac{2b+1}3\le k < b:$ The left side is at least $4$ and the right side is at most $4$.
In case 4, note that since $b$ is odd and $k$ is an integer, $\dfrac b2\le k$ implies that $\dfrac{b+1}2\le k$.
As you can see, the inequality holds in every case. I also only used the fact that $b$ is odd (and greater than $1$), not that it is prime, so I believe it holds for all odd $b>1$ and $a\ge b$.
|
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|
Prove this inequality $25ab+25a+10b\le38$ let $a,b>0$,and such $a^2+b^2=1$,show that
$$25ab+25a+10b\le38$$
Now I have found this inequality $"="$,if and only if $a=\dfrac{4}{5},b=\dfrac{3}{5}$
then How to prove this inequality by AM-GM or other ?
|
From the given conditions we can write, $a=\sin \theta, b=\cos \theta,\ \theta\in (0,\pi/2)$. Then, the objective function becomes $$f(\theta)=25/2\sin 2\theta+25\sin \theta+10\cos\theta\\ f'(\theta)=25\cos 2\theta+25\cos\theta-10\sin \theta$$Equate this to $0$ to get a solution of $\theta$, I believe it pertains to solving a cubic equation $100\cos^3\theta-71\cos\theta+21=0$ which results in three real solutions one of which, that maximizes the function, is $\cos \theta=\frac{3}{5},\implies \sin \theta=\frac{4}{5},\ \sin 2\theta=\frac{24}{25}\implies f(\theta)\le 38,\ \forall \theta\in (0,\pi/2)$.
|
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|
Proving $\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}} < 3$ for $n\geq 1$ by induction How can I prove by induction that $\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}} < 3$ for $n\geq 1$? My guess is that there must be another form to express the sum of nested square roots, but I don't know how to find it.
|
$$(1) \quad x=k+\sqrt x$$
$$x=\sqrt {k+\sqrt x}$$
continuing the recursion...
$$x=\sqrt {k+\sqrt {k+\sqrt {k+...}}}$$
Thus (1) is the equivalent expression.
Solve for x with $k=1$
$$x=\phi$$
Thus, x equals the golden ratio. Multiply the expression by c...
$$c \cdot x=c \cdot \sqrt {k+\sqrt {k+\sqrt {k+...}}}$$
$$c \cdot x=\sqrt {c^2 \cdot k+\sqrt {c^4 \cdot k+\sqrt {c^8 \cdot k+...}}}$$
Choose c to equal $2^{1/4}$ (This is where you use induction to prove this works)
Prove $$2^{2^n/4} \ge n$$
For $n=1 \ $, $ \ \sqrt 2 \gt 1$ so we have a base case
Inductive step, assume the above holds for $n=k$
Let, $n=k+1$
$$2^{2^k/4} \ge k$$
$$2 \cdot 2^{2^k/4} \ge 2 \cdot k$$
$$ 2^{2 \cdot 2^{k}/4} \ge 2 \cdot k$$
$$2^{2^{k+1}/4} \ge k+1$$
assuming k is greater than 1, thus the above holds for all n.
$$2^{1/4} \cdot \phi=\sqrt {\sqrt 2+\sqrt {2+\sqrt { 4+\sqrt {16+...}}}}$$
Thus the inequality for the radical is given by
$$\phi \le S \le 2^{1/4} \cdot \phi$$
$$1.618... \le S \le 1.924...$$
I personally guess the radical to be approximately equal to $1.771$...
|
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|
Can the given transformation possible for given determinant?
In forth step $(x-1)(x-2)$ is obtained by applying transformation R$1 \frac{1}{(x+1)}$ and R$2 \frac{1}{(x+2)}$.
But we get value of $x = -1$ or $ x = -2$ so $\frac{1}{(x+1)}$ and $\frac{1}{(x+2)}$ will be undefined because value of x can be -1 and -2.
So my question is can we apply R1i and R2i?
|
In Step four, $(x+1) , (x+2)$ is obtained not by applying transformation $R_{1\frac{1}{x+1}}$ and $R_{2\frac{1}{x+2}}$. Since determinant function is linear in each row, we have
\begin{align}det
\begin{bmatrix}
c.a_{1,1} & c.a_{1,2} & c.a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}
\end{bmatrix} = c .det
\begin{bmatrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}
\end{bmatrix}
\end{align}
Above identity is true for every row not just first row.
Hence we have the step four using the linearity of determinant in first and third row.
|
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|
Solve the recurrence of the alternating sum $R_n=R_{n-1}+(-1)^{n}(n+1)^{2}$ I have been trying to solve this recurrence for a few hours, but I haven't been able to find the solution yet:
$R_0=1$
$R_n=R_{n-1}+(-1)^{n}*(n+1)^{2}$.
I have been trying to substitute $T_n=(-1)^{n}*R_n$ and then solving for $T_n$ and got the sum:
$1+3+...+\frac{n(n+1)}{2}$
but this sum $(-1)^{n}\frac{n(n+1)(n+2)}{6}$ didn't give the generalized form that would give the terms of the recurrence.
Additionally, I have been trying to substitute $n=2*a$:
$\sum_{k=1}^a (-1)^{2k}*(2k+1)^{2}+ \sum_{k=1}^a (-1)^{2k-1}*(2k)^2$
and I got $\frac{n*(n+3)}{2}$, but it doesn't seem to be right either.
Please help me and if you can figure out, please tell me what I did wrong.
Edited: I have added the initial values.
|
@hypergeometric's solution is my favorite, but another (very general) way of solving this recurrence is by a so-called generating function. Let us define $f(x) = \sum_{n=0}^\infty R_n x^n$. Then,
$$ R_n = R_{n-1} + (-1)^n(n+1)^2 \implies R_nx^n = x R_{n-1} x^{n-1} + (-1)^n (n+1)^2 x^n$$
Summing both sides from $n=1$ to $\infty$, we obtain
$$ f(x)-1 = xf(x) + \sum_{n=1}^\infty (-1)^n (n+1)^2 x^n $$
We can now evaluate the remaining sum by noting
$$ \frac{d^2}{dx^2} \frac{1}{1-x} = \sum_{n=2}^\infty n(n-1)x^{n-2} = \sum_{n=0}^\infty (n+2)(n+1) x^n = \sum_{n=0}^\infty (n+1)^2 x^n + \sum_{n=0}^\infty (n+1)x^n$$
Thus (changing the limits of summation from $n=0$ to $n=1$),
$$ \frac{2}{(1-x)^3} - \frac{1}{(1-x)^2} - 1 = \sum_{n=1}^\infty (n+1)^2 x^n $$
This (substituting $-x$ for $x$) gives us
$$ (1-x)f(x) = \frac{2}{(1+x)^3} - \frac{1}{(1+x)^2} \implies f(x) = \frac{1}{(1+x)^3} = \sum_{n=0}^\infty \underbrace{\frac{(-1)^n}{2}(n+1)(n+2)}_{R_n}x^n.$$
The trick was that we found $f(x)$ in closed form, then re-expanded as a series, since it was of a well-known form. The coefficients of that series are, by construction, $R_n$.
|
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|
Proof of $\sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n}$ Prove that for $n\geq 2, \: \sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n} $
I used induction and I compared the LHS and the RHS but I'm getting an incorrect inequality
|
Meat of induction step:
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{i^2} &= \frac{1}{(k+1)^2}+\sum_{i=1}^k\frac{1}{i^2}\\[1em]
&< \frac{1}{(k+1)^2}+2-\frac{1}{k}\tag{by ind. hyp}\\[1em]
&= 2-\frac{1}{k+1}\left(\frac{k+1}{k}-\frac{1}{k+1}\right)\\[1em]
&= 2-\frac{1}{k+1}\left(\frac{k^2-k}{k(k+1)}\right)\\[1em]
&< 2-\frac{1}{k+1}\tag{since $k\geq 2, k^2-k>0$}
\end{align}
|
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|
Find $\sup_{x\in[0,1]} \frac{x}{x^2+n^2+1}$ We have $f_n:[0,1]\to \mathbb{R},\:f_n(x)=\frac{x}{x^2+n^2+1}$ and we need to prove that is uniform convergence using formula:
$\lim _{n\to \infty } \sup_{x\in[0,1]} |f_n(x)-f(x)| =0$
First step I prove that the sequence $f_n$ converges pointwise to $f(x)=0$. After it: $|f_n(x)-f(x)|=\frac{x}{x^2+n^2+1}$ and $(\frac{x}{x^2+n^2+1})'=\frac{-x^2+n^2+1}{(x^2+n^2+1)^2}$ but I don't understand what will help us to find supremum, in this case.
I need some help to find $\sup_{x\in[0,1]} \frac{x}{x^2+n^2+1}$.
P.S: The author says that $\sup_{x\in[0,1]} \frac{x}{x^2+n^2+1}$ is equal with $\frac1{2n}$, but I think he's wrong, I'm not sure.
|
$f_n(x)$ is a positive function over $(0,1)$ and by the AM-GM inequality
$$ f_n(x)=\frac{x}{x^2+(n^2+1)}=\frac{1}{x+\frac{n^2+1}{x}}\leq\frac{1}{2\sqrt{n^2+1}}\tag{1} $$
with equality attained in $x=\sqrt{n^2+1}$. It follows that $f_n$ is an increasing function over $[0,1]$ and its supremum is simply given by $f_n(1)=\frac{1}{n^2+2}$.
|
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|
Find Primitive Root for Polynomial Field Can someone help me get started on the problem below:
Recall that $\mathbf{F}_{p^k}$ can be realized as $\mathbf{F}_p[x]/P(x)
\cdot \mathbf{F}_p[x]$ where $P(x)$ is a polynomial of degree $k$ with
coefficients in $\mathbf{F}_p$ which is irreducible.
Problem Find primitive roots for the fields $\mathbf{F}_8
\cong \mathbf{F}_2[x]/(x^3+x+1) \cdot \mathbf{F}[x]$ (with elements $a+bx+cx^2$
for $a,b,c \in \mathbf{F}_2$) and for $\mathbf{F}_9 \cong \mathbf{F}_3[x]/(x^2+1)$ (with
elements $a+bx$ for $a,b \in \mathbf{F}_3$). To do it, you just need to
show that some polynomial has multiplicative order equal to $7$ in
the first case and $8$ in the second case.
|
In the quotient $\mathbf{F}_8 \cong \mathbf{F}_2/(x^3+x+1)$ you have the relation
$$x^3+x+1=0 \iff x^3=-x-1=x+1,$$
since over $\mathbf{F}_2, -1=1$. You can use this relation to perform multiplications. In this case $\mathbf{F}_8^{\times} \cong \mathbf{Z}/7\mathbf{Z}$. By Lagrange the order of an element divides the order of the group, so it divides $7$, which is prime in $\mathbf{Z}$, therefore it is either $1$ or $7$, but the only element which has order $1$ is the unit $1$, so every other element has order $7$.
We can prove this directly, so that you get comfortable with how to use the relation above. Let us take $x \in \mathbf{F}_8$. (Again, you can take whatever element you like, so, if you want, just try it) We show that $\langle x \rangle =\mathbf{F}_8^{\times}$.
$x^1 = x;$
$x^2 =x^2;$
$x^3 = x+1;$
$x^4 = x\cdot x^3 = x(x+1)=x^2+x;$
$x^5 = x^2\cdot x^3 = x^2(x+1)=x^3+x^2=x+1+x^2;$
$x^6 = x^3\cdot x^3 = (x+1)^2 = x^2+2x+1=x^2+1,$ since over $\mathbf{F}_2, 2=0$;
$x^7 = x^6 \cdot x = (x^2+1)x = x^3+x = x+1+x=1.$
I let you do the other case. Just notice that the relation it holds this time is $x^2+1=0$...
If you have problems with it just ask :-).
|
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Finding range of $||x| - |y||$ for the given conditions.
If $ z = x + iy$ and $ x^2 + y^2 = 16 $ , then the range of $||x|-|y||$ is...?
This is what I've tried yet:
Suppose $x = a\cos \theta$ and $y = b\sin \theta$, then we've :
$$\begin{align}
x^2 + y^2 = 16 \implies & a^2\cos^2 \theta + b^2\sin^2 \theta= 16 \\
\implies & a^2 + \sin^2 \theta(b^2 - a^2) = 16 \\
\implies & \sin^2 \theta = \cfrac{16 - a^2}{b^2 - a^2} \tag{1}
\end{align}$$
Not sure from where to go on from here. Putting the same values in $ z = x+iy$ and then squaring both sides doesn't give anything fruitful too.
|
Hint:-
Put $x=4\sin\theta$ and $y=4\cos \theta$.
Solution
Then, $$||x|-|y||=4||\sin\theta|-|\cos \theta||$$Now, $$0\le|\sin\theta|\le 1$$and, $$-1\le-|\cos\theta|\le0$$which gives, $$||\sin\theta|-|\cos \theta||\le1\implies 4||\sin\theta|-|\cos \theta||\le4$$
|
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|
Minimise $ab+bc+ac$ Let $a,b,c \in \mathbb R$, and $a^2+b^2+c^2=1$
How can I calculate the minimum value of $ab+bc+ac$? (i.e. most negative)
I've tried using the fact that $(a-b)^2+(b-c)^2+(a-c)^2 \ge 0$ but this gives an inequality in the wrong direction.
|
Note that $f(a,b,c) = ab+bc+ac = {1 \over 2} ((a+b+c)^2-(a^2+b^2+c^2))$.
Hence $\min \{ f(a,b,c) | a^2+b^2+c^2=1 \} = {1 \over 2} \min \{ (a+b+c)^2-1 | a^2+b^2+c^2=1 \}$.
Note: See Barry's comment below for a simpler alternative to the following.
One way that doesn't use Lagrange multipliers is to change basis.
Let $u_1 = {1 \over \sqrt{3}} (1,1,1)$ and let $u_2,u_3$ be such that $u_k$ form an orthonormal basis. In this basis the problem becomes
${1 \over 2}\min \{ 3x^2-1 | x^2+y^2+z^2 = 1\} $ and, by inspection, we see that
the minimum is $-{1 \over 2}$.
|
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|
How can I integrate $\int {dx \over \sqrt{3^2+x^2}} $ using Trigonometric Substitution? $$\int {dx \over \sqrt{9+x^2}} = \int {dx \over \sqrt{3^2+x^2}} $$
$$ x =3\tan\theta$$
$$dx = 3\sec^2\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2 + 3^2\tan^2\theta}} d\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2(1+\tan^2\theta)}} d\theta$$
$$\int {3\sec^2\theta \over 3\sec\theta} d\theta = \int \sec\theta$$
$$\ln|\sec\theta + \tan\theta| + C $$
$$\ln\left({\sqrt{9+x^2} \over 3} + {x \over 3}\right)$$
My book says the answer should just be:
$$\ln\left({\sqrt{9+x^2}} + {x}\right) $$
I'm wondering where I went wrong with this?
|
It should be a printing mistake. Your answer is correct
|
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|
The complex roots of a biquadratic polynom In my recent post I have a problem with the following function: $x^4-4x^2+16$, and what I need is to find the complex roots.
Here is my answer:
First step, I make the substitution $x^2=y$ which involving $y^2-4y^2+16$, with $x^2=2\pm i\sqrt{12}$. Therfore:
$x^4-4x^2+16=(x^2-2-i\sqrt{12})(x^2-2+i\sqrt{12})
=(x+\sqrt{2+i\sqrt{12}})(x+\sqrt{2-i\sqrt{12}})(x-i\sqrt{2+i\sqrt{12}})(x-\sqrt{2-i\sqrt{12}})$
factorized irreducible over $\mathbb{R}\subset\mathbb{C}$.
Here is author's answer:
$x{_1,_2,_3,_4}=\pm(\sqrt{3}\pm i)$.
I realize the problem is at factorization: $x^4-4x^2+16=(x^2+bx+c)(x^2+dx+e)$.
If I pair the complex roots from author's answer I'll obtain:
$x^4-4x^2+16=(x+\sqrt{3}+i)(x+\sqrt{3}-i)(x-\sqrt{3}-i)(x-\sqrt{3}+i)=(x^2+\sqrt{12}x+4)(x^2-\sqrt{12}x+4)$
factorized irreducible over $\mathbb{Q}\subset\mathbb{R}$, but reducible over $\mathbb{C}$.
My final question: Why $x_1{,_2,_3,_4}=\pm\sqrt{2\pm i\sqrt{12}}$ isn't enough to make the assessment that it is the final answer ? How can I get author's answer $x{_1,_2,_3,_4}=\pm(\sqrt{3}\pm i)$ begining from my complex roots found ? Why we obtain 2 different form of complex roots, who's belong in the same field $\mathbb{C}$ when we change factorization ?
|
Hint:
with your substitution, the solutions that you have are:
$$
x^2=2(1\pm i\sqrt{3})=4 e^{\pm i \frac{\pi}{3}}
$$
and this gives immediately the autor's answer.
added:
The complex number $a+ib=2+2i\sqrt{3} \rightarrow $ can be write in polar form $\rho e^{i\theta}$ with:
$$
\rho=\sqrt{a^2+b^2}=\sqrt{4+12}=4
$$
$$
\theta= \arctan\left(\dfrac{b}{a}\right)=\arctan\left(\dfrac{2\sqrt{3}}{2}\right)= \dfrac{\pi}{3}
$$
And if $x^2=4 e^{\pm i \frac{\pi}{3}}$ we have $x=\pm 2 e^{\pm i \frac{\pi}{6}}$, that, in binomial form, are the solutions.
|
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|
Differentiation method for evaluating $ \sum_{n=1}^\infty \frac{n^2}{3^n} $ I evaluated the following infinite sum (the original and broader question regarding this sum can be found at Evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n} $).
$$ \sum_{n=1}^\infty \frac{n^2}{3^n} $$
However, I'm getting the feeling that I made some mistake(s) during my evaluations. My particular concern regards the changes made in the sum's index. The following is what I did.
I first defined a power series as the function f.
$$ \sum_{n=1}^\infty \frac{n^2}{3^n} = \sum_{n=1}^\infty n^2 (\frac{1}{3})^n = f(\frac{1}{3}) \Rightarrow f(x) = \sum_{n=1}^\infty n^2 x^n $$
I then attempted to manipulate the sum in order to transform it through the geometric series. This is where I'm quite unsure whether I did everything correctly. One of the things I did here looks wrong to me, but somehow, I still ended up with the correct answer (which is 3/2).
$$
\begin{align*}
f(x) &= \sum_{n=1}^\infty n^2 x^n
\\ &= x \sum_{n=1}^\infty n^2 x^{n-1}
\\ &= x \frac{d}{dx} ( \sum_{n=0}^\infty n x^n )
\\ &= x \frac{d}{dx} ( x \sum_{n=0}^\infty n x^{n-1} )
\\ &= x \frac{d}{dx} ( x \frac{d}{dx} ( \sum_{n=-1}^\infty x^n ) )
\\ &= x \frac{d}{dx} ( x \frac{d}{dx} ( \frac{1}{1-x} ) )
\\ &= x \frac{d}{dx} ( x \frac{1}{(1-x)^2} )
\\ &= x \frac{d}{dx} ( \frac{x}{(1-x)^2} )
\\ &= x \frac{1+x}{(1-x)^3}
\\ &= \frac{x(1+x)}{(1-x)^3}
\\ &\Rightarrow f(\frac{1}{3}) = \frac{3}{2}
\end{align*}
$$
The main concern of mine is the transition step to the closed-form geometric series. Of course, the proper equation for a geometric series is this:
$$ \sum_{n=0}^\infty x^n = \frac{1}{1-x} $$
However, what I did is this:
$$ \sum_{n=-1}^\infty x^n = \frac{1}{1-x} $$
The difference here is that the starting index is -1 instead of 0. This makes my translation of the sum incorrect. And yet, I still get the correct answer. On the other hand, I've tried the correct(?) form of an infinite geometric series that starts at n=-1:
$$ \sum_{n=-1}^\infty x^n = \frac{\frac{1}{x}}{1-x} = \frac{1}{x(1-x)} $$
However, this yields an incorrect answer.
I'm guessing that the final index I should've had for the sum was n=0 instead of n=-1. I'm guessing I did something wrong with the index shifts caused by the derivatives? Either way, I'm not seeing it.
I do realize that there are other ways to go about evaluating this sum, but I'd really like to understand this derivative method.
|
You could introduce a step in which you write
$$
\begin{align}
f(x) & = \cdots \\
& = x \frac{d}{dx} \left( x \sum_{n=0}^\infty nx^{n-1} \right) \\
& = \color{red}{
x \frac{d}{dx} \left( x \sum_{n=1}^\infty nx^{n-1} \right)} \\
& = x \frac{d}{dx} \left( x \frac{d}{dx}
\left( \sum_{n=0}^\infty x^n \right) \right) \\
& = \cdots
\end{align}
$$
since $nx^{n-1} = 0$ when $n = 0$. That should resolve matters, although it isn't in fact necessary, because$\ldots$
ETA: Actually, your line is in error. You write
$$
\begin{align}
f(x) & = \cdots \\
& = x \frac{d}{dx} \left( x \sum_{n=0}^\infty nx^{n-1} \right) \\
& = x \frac{d}{dx} \left( x \frac{d}{dx}
\left( \sum_{n=-1}^\infty x^n \right) \right) \\
& = \cdots
\end{align}
$$
but the index limit should have remained $n = 0$, since
$$
\begin{align}
f(x) & = \cdots \\
& = x \frac{d}{dx} \left( x \sum_{n=0}^\infty nx^{n-1} \right) \\
& = \color{red}{
x \frac{d}{dx} \left( x \sum_{n=0}^\infty \frac{d}{dx} x^n \right)} \\
& = x \frac{d}{dx} \left( x \frac{d}{dx}
\left( \sum_{n=0}^\infty x^n \right) \right) \\
& = \cdots
\end{align}
$$
|
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|
Compute limit using Taylor's expansion Using Taylor’s expansion, prove that the following limit exists and compute it.
$$\lim_{x \to 0}\left(\frac {x^2}{\frac {1}{1-x} - e^x}\right)$$
In this if I am using the taylor series expansion of $e^x$ then denominator has some value as $$ 1 + (1 + x + x^2 + x^3 + \ldots) + x (1 + x + x^2 + x^3 + \ldots) $$
I am not getting how to proceed further with this.
|
Hint: First we have
$$\frac{1}{1 - x} - e^x = \sum_{n=0}^\infty x^n - \sum_{n=0}^\infty \frac{x^n}{n!} =\sum_{n=0}^\infty x^n \Big(1 - \frac{1}{n!}\Big) = \frac{x^2}{2} + \frac{5x^3}{6} + O (x^4) $$
Then
$$\lim_{x \to 0}\frac{x^2}{\frac{x^2}{2} + \frac{5x^3}{6} + O (x^4)}$$
Can you take it from here?
|
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|
Solving a recurrence relation using generating functions My recurrence relation is
D(n) = D(n - 1) + D(n - 2) + 5(n - 1);
with the initial conditions D(2),D(3) being 6, 17 respectively.
The generating function G(z) for the sequence D(n) is given
I don't know how i got this. Please can anyone give explanation for this?
Thank you.
|
Define the generating function:
$$
G(z) = \sum_{n \ge 0} D(n) z^n
$$
Take your recurrence written as:
$$
D(n + 2) = D(n + 1) + D(n) + 5 (n + 1)
$$
Multiply by $z^n$, sum over $n \ge 0$ and recognize some sums:
$$
\frac{G(z) - D(0) - D(1) z}{z^2}
= \frac{D(z) - D(0)}{z} + G(z) + \frac{5}{(1 - z)^2}
$$
Solve for $G(z)$, split into partial fractions:
$\begin{align}
G(z)
&= \frac{D(0) + (D(1) - 3 D(0)) z
- (2 D(1) - 3 D(0) - 5) z^2
+ (D(1) - D(0)) z^3}
{(1 - z)^2 (1 - z - z^2)} \\
&= \frac{(D(0) + 10) + (D(1)- D(0) + 5) z}{1 - z - z^2}
- \frac{5}{1 - z} - \frac{5}{(1 - z)^2}
\end{align}$
Now we know that the Fibonacci numbers $F_n$ are defined by:
$$
F_{n + 2} = F_{n + 1} + F_n \qquad F_0 = 0, F_1 = 1
$$
with generating function:
$$
F(z) = \sum_{n \ge 0} F_n z^n = \frac{z}{1 - z - z^2}
$$
so that $1 / (1 - z - z^2)$ corresponds to the sequence of the $F_{n + 1}$, and so:
$$
D(n)
= (D(0) + 10) F_{n + 1}
+ (D(1) - D(0) + 5) F_n
- 5 - 5 (n + 1)
$$
Plug in the known values of $D(2)$ and $D(3)$ to get a system of equations for the missing $D(0)$ and $D(1)$, and you are done.
|
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|
Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$ Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$. I put $(2+2x)^3=3-2\cdot3^{x+1}$. But I don't know how to go on.
|
$$3-2\cdot 3^{x+1} = (3^{x+1})^2$$
Set $z = 3^{x+1}$:
$$z^2 + 2\cdot z-3 = 0$$
Solve quadratic equation:
$$z = -1 \pm \sqrt{1 + 3} = -1 \pm 2$$
so $z \in \{-3,1\}$.
Compute $x$ from
$$x = \log_3(z) - 1.$$
For a real solution, you have to pick $z=1$, so
$$x = \log_3(1) - 1 = 0 - 1 = -1.$$
|
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|
Prove non-existance of limit: $f(x,y) = \frac{xy\sin(\frac{x}{y})}{x^2 + |y|^3}$ I need to prove that $f(x,y) = \frac{xy^2\sin(\frac{x}{y})}{x^2 + |y|^3}$ does not tend to $0$ when $(x,y)$ approaches $(0,0)$.
In order to do so, I would need to find some direction $\alpha$ such that $f(\alpha(t))$ approaches to some value $L \neq 0$ as $(x,y)$ approaches the origin.
The problem is that I can't seem to find that direction. What could I try?
I found $$f(x^\frac{1}{2}, x^\frac{1}{3}) = \frac{x^\frac{7}{6} \sin(x^\frac{-1}{6})}{2x} = \frac{x^\frac{1}{6} \sin(x^\frac{-1}{6})}{2} = \frac{\sin(x^\frac{-1}{6})}{2 x^\frac{-1}{6}} \rightarrow \frac{1}{2}$$
Is this correct?
|
Used polar coordinates $x=r\cos\phi, y=r\sin\phi$ and compute the limit $r \rightarrow 0$. The result is
\begin{align}
\lim_{r\rightarrow 0} \frac{r^2\cos\phi\sin\phi \cdot \sin(\cot\phi)}{r^2 \cos^2\phi + |r\sin\phi|^3}
&=
\lim_{r\rightarrow 0} \frac{\cos\phi\sin\phi \cdot \sin(\cot\phi)}{\cos^2\phi + r|\sin^3\phi|}
\\ &=
\frac{\cos\phi\sin\phi \cdot \sin(\cot\phi)}{\cos^2\phi}
\\ &= \tan\phi \cdot \sin(\cot\phi)
\end{align}
which is a nonconstant function of $\phi$, thus your limit does not exist.
EDIT: The answer of Mann is easier and more straightforward though.
|
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|
Intuitively understanding $\sum_{i=1}^ni={n+1\choose2}$ It's straightforward to show that
$$\sum_{i=1}^ni=\frac{n(n+1)}{2}={n+1\choose2}$$
but intuitively, this is hard to grasp. Should I understand this to be coincidence? Why does the sum of the first $n$ natural numbers count the number of ways I can choose a pair out of $n+1$ objects? What's the intuition behind this?
|
The intuition is that for the pairs can be listed in the following way.
$$\begin{array}{ccccccc}
1,2 & & & & & & \\
1,3 & 2,3 & & & & & \\
1,4 & 2,4 & 3,4 & & & & \\
1,5 & 2,5 & 3,5 & 4,5 & & & \\
1,6 & 2,6 & 3,6 & 4,6 & 5,6 & & \\
1,\vdots & 2,\vdots & 3,\vdots & 4,\vdots & 5,\vdots &\ddots & \\
1,n+1 & 2,n+1 & 3,n+1 & 4,n+1 & 5,n+1 & \cdots & n,n+1 \\
\end{array}$$
Notice that each row has length $i$ for $i=1,\ldots,n$ since the number of pairs with maximum element $i+1$ is $i$. Therefore the total number of pairs, which is $\binom{n+1}{2}$ is $\displaystyle \sum_{i=1}^n i$.
|
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|
Generating function of $a_{n} = n$ I want to find the closed form of this $\sum\limits_{n=0}^{\infty} {nx^{n}}$Is the following correct?
$(\sum\limits_{n=0}^{\infty} x^n)^{'}= (1)^{'}+(x)^{'}+(x^2)^{'}+(x^3)^{'}+...=$
$0 + 1 +2x + 3x^2 + ...=$
$=(\sum\limits_{n=1}^{\infty} {nx^{n-1}})=(\sum\limits_{k=0}^{\infty} {(k-1)x^{k}})= (\sum\limits_{k=0}^{\infty} {kx^{k}}) - (\sum\limits_{k=0}^{\infty} {x^{k}})$ $,(1)$
But this $\sum\limits_{k=0}^{\infty} {x^{k}} = \frac{1}{1-x}$ and $(\sum\limits_{n=0}^{\infty} x^n)^{'} = (\frac{1}{1-x})^{'}=\frac{1}{(1-x)^2}$
So from $(1)$ we have $(\sum\limits_{n=0}^{\infty} {x^{n}})^{'}=(\sum\limits_{k=0}^{\infty} {kx^{k}}) - (\sum\limits_{k=0}^{\infty} {x^{k}})\Rightarrow\frac{1}{(1-x)^2}=(\sum\limits_{n=0}^{\infty} {nx^{n}})-\frac{1}{1-x}$
According to my notes it should be equal $\frac{x}{(1-x)^{2}}$ thoough, but I can't find my mistake.
|
What you did is correct except that
$$
\sum\limits_{n=1}^{\infty} {nx^{n-1}}=\sum\limits_{k=0}^{\infty} {(k+1)x^k}
$$ and we are led to the correct result
$$
\sum\limits_{n=1}^{\infty} {nx^{n-1}}=\frac{1}{(1-x)^2}.
$$ or equivalently, multiplying by $x$:
$$
\sum\limits_{n=1}^{\infty} {nx^{n}}=\frac{x}{(1-x)^2}.
$$
|
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|
Finding roots of 5 th degree taylor expansion of $e^x$ I need to find the roots [actually their count and sign] of :
$$
{\mathcal T}(e^x,5)=\frac{x^5}{5!}+\frac{x^4}{4!}+\frac{x^3}{3!}+\frac{x^2}{2!}+x+1$$
*
*It[Root] is clearly negative.
*It should have odd number of roots.
*Either One or Three or Five.
*It is easy to find the derivatives, but only the third derivative, a quadratic can be easily analyzed.
Any help?
Basics Only.
|
Consider the derivative
$$\begin{aligned}
\frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + x + 1 &= \frac{1}{4!}(x^4 + 4x^3 + 4x^2) + \frac{1}{12}x^2 + \left(\frac{x^2}{4} + x + 1\right)\\
&= \frac{x^2}{24}(x+2)^2 + \frac{x^2}{12} + \left(\frac{x}{2}+1\right)^2\\
&> 0.
\end{aligned}$$
So the fifth-order Taylor polynomial is strictly increasing, hence has exactly one real zero.
|
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|
Counterclockwise rotation matrix If I take the basis $(\vec{e_x},\vec{e_y})$ and make a rotation counterclockwise of angle $\theta$, I end up with two new vectors $(\vec{u},\vec{v})$ such that :
$\vec{u} = \cos\theta \vec{e_x} + \sin\theta \vec{e_y}$
$\vec{v} = \cos\theta \vec{e_x} - \sin\theta \vec{e_y}$
so
\begin{equation}
\left( \begin{array}{ccc}
\vec{u} \\
\vec{v}\end{array} \right)
= \left( \begin{array}{ccc}
\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta\end{array} \right)
\left( \begin{array}{ccc}
\vec{e_x} \\
\vec{e_y}\end{array} \right)
\end{equation}
I don't understand why the counterclockwise rotation is defined as :
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
EDIT:
When I look at my picture, it looks like a counterclockwise rotation...
|
You can also do it in a more algebraic way. Since after rotation ($(x, y)$ is rotated to $(x', y')$), the length of the vector doesn't change, which means $n = \sqrt{x'^2 + y'^2} = \sqrt{x^2 + y^2}$ (see in figure attached).
Therefore we can get the following equation:
\begin{aligned}
y' & = n \cdot \sin(\theta + \alpha) & (1)\\
y & = n \cdot \sin \alpha & (2)
\end{aligned}
\begin{aligned}
x' & = n \cdot \cos(\theta + \alpha) & (3) \\
x & = n \cdot \cos \alpha & (4)
\end{aligned}
Then use the trigonometric identities to expand (1) and (3):
\begin{aligned}
y' & = n \cdot (\sin \theta \cos \alpha + \cos \theta \sin \alpha) & (5) \\
x' & = n \cdot (\cos \theta \cos \alpha - \sin \theta \sin \alpha) & (6)
\end{aligned}
By substituting (2) and (4) into (5) and (6), we can get:
\begin{aligned}
y' & = x \cdot \sin \theta + y \cdot \cos \theta \\
x' & = x \cdot \cos \theta - y \cdot \sin \theta
\end{aligned}
From here we can easily see:
$$\begin{bmatrix}x'\\y'\end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}$$
|
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|
What exactly IS a square root? It's come to my attention that I don't actually understand what a square root really is (the operation). The only way I know of to take square roots (or nth root, for that matter) it to know the answer! Obviously square root can be rewritten as $x^{1/2}$ , but how does one actually multiply something by itself half a time?
How do calculators perform the operation?
|
I don't believe any computer actually does it this way, but you can compute a continued fraction to get an exact representation of a square root. For example, since $10^2$ $<$ $111$ $<$ $11^2$:
\begin{eqnarray}
\sqrt{111}
&=& 10 + (\sqrt{111} - 10) \\
&=& 10 + \frac{11}{\sqrt{111} + 10} = 10 + \frac{1}{\frac{\sqrt{111} + 10}{11}} \\
\frac{\sqrt{111} + 10}{11} &=& 1 + \frac{\sqrt{111} - 1}{11} \\
&=& 1 + \frac{10}{\sqrt{111} + 1} = 1 + \frac{1}{\frac{\sqrt{111} + 1}{10}} \\
\frac{\sqrt{111} + 1}{10} &=& 1 + \frac{\sqrt{111} - 9}{10} \\
&=& 1 + \frac{3}{\sqrt{111} + 9} = 1 + \frac{1}{\frac{\sqrt{111} + 9}{3}} \\
\frac{\sqrt{111} + 9}{3} &=& 6 + \frac{1}{\frac{\sqrt{111} + 9}{10}} \\
\frac{\sqrt{111} + 9}{10} &=& 1 + \frac{1}{\frac{\sqrt{111} + 1}{11}} \\
\frac{\sqrt{111} + 1}{11} &=& 1 + \frac{1}{\sqrt{111} + 10} \\
\sqrt{111} + 10 &=& 20 + \frac{1}{\frac{\sqrt{111} + 10}{11}}
\end{eqnarray}
This yields the continued fraction $[10, \dot1, 1, 6, 1, 1, \dot{20}]$, and it looks like:
$$
\sqrt{111} = 10 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{6 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{20 + \cfrac{1}{...}}}}}}}
$$
|
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|
Solving $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $. I have the equation $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $.
I tried to square both sides, but then I got a more difficult equation:
$$
2 x - 11 + 2 \sqrt{x^{2} - 11 x - 28} = 1.
$$
Can someone tell me what I should do next?
|
$$\sqrt{x-4}+\sqrt{x-7}=1<=>$$
$$-11+2\sqrt{(x-7)(x-4)}=12-2x<=>$$
$$4(x-7)(x-4)=(12-2x)^2<=>$$
$$4x^2-44x+112=4x^2-48x+144<=>$$
$$4x-32=0<=>$$
$$4(x-8)=0<=>$$
$$x-8=0<=>$$
$$x=8$$
BUT THIS SOLUTION IS INCORRECT SO THERE ARE NO SOLUTIONS!!!!!
|
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|
What's wrong?: Find the infinite sum $S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$ The answer I got by hand is not the same to the one I found using a spreadsheet.
$\displaystyle S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$
$\displaystyle \frac{1}{4}S = \hspace{8.5pt} \frac{1}{4} + \frac{3}{16} + \frac{7}{64} + \frac{15}{256} + \ldots$
$\displaystyle \frac{3}{4}S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \qquad \leftarrow S- \frac{1}{4}S$
For the Infinite Sum on the RHS $\displaystyle \left(S = \frac{a}{1-r}\right)$:
$\displaystyle a = 1$
$\displaystyle r = \frac{1}{2}$
Then
$\displaystyle \frac{3}{4}S = \frac{1}{1-\frac{1}{2}}$
$\displaystyle \frac{3}{4}S = 2$
$\displaystyle S = \frac{8}{3}$
Using Excel the answer is $\displaystyle \frac{5}{3}$, but I don't know where is the issue.
Thanks!!
|
Hint: Notice that
$$
S = \sum_{i=0}^\infty \frac{2^{i+1}-1}{2^{2i}}
$$
and distribute.
|
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|
Limit and Integral problem work verification-2 I have to calculate the following:
$$\large\lim_{x \to \infty}\left(\frac {\displaystyle\int\limits_{x^{2}}^{2x}t^{4}e^{t^{2}}dt}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)$$
My attempt:
Let $F(x)=\displaystyle\int\limits_0^xt^4e^{t^2}dt$. Then,
$$\large\lim_{x\to\infty}\left(\frac{\displaystyle\int\limits_{x^{2}}^{2x}t^{4}e^{t^{2}}dt}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)=\lim_{x \to \infty}\left(\frac {F(2x) - F(x^2)}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)$$
Applying L'Hôpital's rule, we have,
$$\large\begin{align}\lim_{x \to \infty}\left(\frac {32x^4e^{4x^2} - 2x^9e^{x^4}}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}}\right) &= \lim_{x \to \infty}(32x^4e^{4x^2-x} - 2x^9e^{x^4-x}) \\&= \lim_{x \to \infty}\bigg(2x^4e^{4x^2-x}(16-x^5e^{x^4-4x^2})\bigg) = -\infty\end{align}$$
Am I right?
|
Hint: I strongly suggest the use of power series. We can write down the power series for $e^w$, substitute $t^2$, and integrate term by term to get a series for the top. The series for the bottom is easy to write down.
Remark: As mentioned in a comment, there is an error in the L'Hospital's Rule calculation. It is fixable.
|
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|
Need a more compact formula This is a part of solution of a programming contest problem
$$\sum_{i=0}^{k} {x-i \choose 2} $$ given $x-i \ge 2$ is always true.
for
$k=1$,$(x-1)^2$
$k=2$, $(x-1)^2+((x-2)*(x-3)/2)$
$k=3$, $(x-1)^2 + (x-3)^2$
and so on.
Is there a reduced form of this?
|
$$\frac{1}{2}\sum_{i=0}^k (x-i)(x-i-1) = \frac{1}{2} \left( \sum_{i=0}^k (x^2-2xi+i^2 + x-i) \right) =$$
$$=\frac{1}{2} \left( \sum_{i=0}^k (x^2+ x) + \sum_{i=0}^k (-2xi-i) + \sum_{i=0}^k i^2\right)= \frac{1}{2} \left( (k+1)(x^2+x) -(2x+1)\left(\frac{k(k+1)}{2} \right) + \frac{k(k+1)(2k+1)}{6}\right)$$
|
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|
Solve the equation $\log_{2} x \log_{3} x = \log_{4} x$ Question:
Solve the equations
a)
$$\log_{2} x + \log_{3} x = \log_{4} x$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
Attempted solution:
The general idea I have been working on is to make them into the same base, combine them and take the inverse to get the solution for x.
a)
$$\log_{2} x + \log_{3} x = \log_{4} x \Leftrightarrow$$
$$\log_{2} x + \frac{\log_{2} x}{\log_{2} 3} = \frac{\log_{2} x}{\log_{2} 4} \Leftrightarrow$$
$$\log_{2} x + \frac{\log_{2} x}{\log_{2} 3} = \frac{\log_{2} x}{2} \Leftrightarrow$$
$$\frac{2 \log_{2} 3 \log_{2}x + 2\log_{2} x - \log_{2}3 \log_{2}x}{2 \log_{2}3} = 0 \Leftrightarrow $$
Moving the denominator over and solving for $\log_{2} x$
$$\frac{\log_{2} x (2\log_{2}3 + 1 - \log_{2}3)}{2\log_{2} 3} = 0 \Leftrightarrow$$
$$\log_{2} x = 0 \Rightarrow x = 2^{0} = 1$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
$$\log_{2} x \frac{\log_{2} x}{\log_{2} 3} - \frac{\log_{2} x}{\log_{2} 4} = 0 \Leftrightarrow$$
$$\frac{4 (\log_{2}x)^2 - \log_{2}3 \log_{2}x}{4 \log_{2}3} = 0 \Leftrightarrow $$
$$4 (\log_{2}x)^2 - \log_{2}3 \log_{2}x = 0 \Leftrightarrow $$
Substituting $t = \log_{2} x$ gives:
$$4t^2 - \log_{2}3t = 0 \Rightarrow$$
$$t^2 - \frac{\log_{2}3t}{4} = 0 \Rightarrow$$
$$t \left(t- \frac{\log_{2}3}{4}\right) = 0$$
$$t_{1} = 0 \Rightarrow x = 1$$
$$t_{2} = \frac{\log_{2}3}{4} \Rightarrow x = 2^{\frac{\log_{2}3}{4}}$$
However, the second solution here should be $\sqrt{3}$, so I must have made a mistake somewhere. Any suggestions?
|
One solution is apparent for (a): $x=1$. There can be no other solutions, since the two sides' derivatives are $\frac{1}{x}\left(\frac{1}{\ln(2)}+\frac{1}{\ln(3)}\right)>\frac{1}{x}$ and $\frac{1}{x}\frac{1}{\ln(4)}<\frac{1}{x}$. This shows us that the left side is always growing at a faster rate than the right side, so at most one solution is possible.
For (b), $x=1$ is also an apparent solution. But it's not the only one. Raising $4$ to both sides leaves $$x^{2\log_3(x)}=x\implies x^{2\log_3(x)-1}=1\quad\text{(since $x\ne0$)}$$
So $\ln(x)\cdot(2\log_3(x)-1)=0$. Either $x=1$ to make the first factor $0$, or $2\log_3(x)-1=0$ to make the second factor $0$. The latter option implies $x=3^{1/2}$. So the only two possible solutions are $x=1$ and $x=\sqrt{3}$, both of which check out to actually be solutions.
|
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|
Find $S=\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{1+n^2}$ How do I find the sum:
$$S=\sum_{n=-\infty}^{\infty} \dfrac{(-1)^n}{1+n^2}$$
I can't solve this can someone help me?
|
$$S=\sum_{n \ \text{even}} \dfrac{1}{1+n^2}-\sum_{n \ \text{odd}} \dfrac{1}{1+n^2}$$
Now, $$\displaystyle\sum_{n \ \text{even}} \dfrac{1}{1+n^2}=\displaystyle\sum_{n=-\infty}^{\infty} \dfrac{1}{1+(2n)^2}=\dfrac{1}{4}\displaystyle\sum_{n=-\infty}^{\infty} \dfrac{1}{1/4+n^2}$$
Also,
$$\begin{align}\sum_{n \ \text{odd}} \dfrac{1}{1+n^2}&=\sum_{n=-\infty}^{\infty}\dfrac{1}{1+n^2}-\sum_{n \ \text{even}} \dfrac{1}{1+n^2}\\ &=\sum_{n=-\infty}^{\infty}\dfrac{1}{1+n^2}-\dfrac{1}{4}\sum_{n=-\infty}^{\infty} \dfrac{1}{1/4+n^2}\end{align}$$
Hence,
$$S=\dfrac{1}{2}\displaystyle\sum_{n=-\infty}^{\infty} \dfrac{1}{1/4+n^2}-\sum_{n=-\infty}^{\infty}\dfrac{1}{1+n^2} \ \ \ \ \ \ \ \ \ (1)$$
Now, consider $f(x)=e^{-a|x|}, \ a>0$. Let $F(k)$ be the Fourier transform of $f$.
$$\begin{align} \therefore \ F(k)&=\int_{-\infty}^{\infty}f(x)e^{-ikx}\mathrm{d}x\\ &= \int_{-\infty}^{\infty}e^{-a|x|}e^{-ikx}\mathrm{d}x\\ &=\dfrac{2a}{a^2+k^2}\end{align}$$
Now according to the Poisson summation formula,
$$\sum_{n=-\infty}^{\infty}f(n)=\sum_{m=-\infty}^{\infty}F(2\pi m)$$
Therefore,
$$\sum_{n=-\infty}^{\infty}e^{-a|n|}=\sum_{m=-\infty}^{\infty}\dfrac{2a}{a^2+(2\pi m)^2}$$
The LHS is just a geometric series. Summing it, we get
$$\begin{align}\sum_{m=-\infty}^{\infty}\dfrac{2a}{a^2+(2\pi m)^2}&=\dfrac{1+e^{-a}}{1-e^{-a}}\\ \sum_{m=-\infty}^{\infty} \dfrac{1}{(a/2\pi)^2+n^2}&=\dfrac{2\pi^2}{a}\dfrac{1+e^{-a}}{1-e^{-a}}\end{align}$$
Put $a=2\pi,\pi/2$ and substitute the results into $(1)$ to obtain
$$\therefore{S=\dfrac{2\pi e^{\pi}}{e^{2\pi}-1}}$$
|
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|
Show that $ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $ The Question reads -
$$ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $$
I tried using the obvious approach by converting $5x , 4x $ and $ 3x$ to either $2x$ or $x$ but all that seemed to do was to further complicate the fraction. Any hints would be much appreciated.
I also tried applying various identities but to no avail.
|
we can use product to sum formula. we have
$$\begin{align}(1- 2\cos 3x)(-\cos 2x - \cos x) &= 2\cos 3x \cos 2x + 2 \cos 3x \cos x - \cos 2x - \cos x\\
&= (\cos 5x + \cos x) + (\cos 4x + \cos 2x) - \cos 2x - \cos x\\
&=\cos 5x+\cos 4x\end{align}$$
dividing it out by $1- 2\cos 3x$ gives the result.
|
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|
Prove that $7$ divides $1 + 2^{(2^n)} + 2^{(2^{n+1})}$ by induction Prove that $7$ divides $1 + 2^{(2^n)} + 2^{(2^{n+1})}$ by induction.
I ran into the above problem. The base case $n=1$ gives $21$ which is divisible by $7$.
Now assume it is true for $n$. Then for $n+1$, we have the expression
$$ 1 + 2^{(2^{n+1})} + 2^{(2^{n+2})}$$
which is equal to
$$1 + 2^{(2^n\cdot 2)} + 2^{(2^n \cdot 2^2)}$$
but I do not really see how it helps.
Any ideas?
|
Consider the polynomial
$$f(x) = 1+x^{2^n} + \left(x^{2^{n}} \right)^2$$
We claim that $f(x) = (1+x+x^2)g(x)$, where $g(x)$ is also a polynomial for all $k \in \mathbb{Z}^+$. This is easy to prove, since $\omega = e^{2\pi i/3}$ and $\omega^2 = e^{4\pi i/3}$ are roots of $1+x+x^2$. We have $1+\omega^{2^n} + \left(\omega^{2^n}\right)^2 = 0$. Hence, $1+x+x^2$ divides $1+x^{2^n} + x^{2^{n+1}}$. Taking $x=2$, we obtain what you want.
|
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|
$ \lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15} $ Can someone show me how is possible to prove that
\begin{equation*}
\lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15}
\end{equation*}
but without Taylor series. One can use L'Hospital rule if necessary. I was not able.
|
Let the desired limit be denoted by $L$.
We have via LHR $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}} = \frac{1}{6}\tag{1}$$ and $$\lim_{x \to 0}\frac{\tan x - x}{x^{3}} = \lim_{x \to 0}\frac{\sec^{2} x - 1}{3x^{2}} = \frac{1}{3}\tag{2}$$ and we also have $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{3}$$ Multiplying the 3 limits above we get
\begin{align}
&\lim_{x \to 0}\frac{(x - \sin x)(\tan x - x)\sin x}{x^{7}} = \frac{1}{18}\notag\\ &\Rightarrow\lim_{x \to 0}\frac{(x\sin x\tan x - x^{2}\sin x - \sin^{2}x\tan x + x\sin^{2}x)}{x^{7}} = \frac{1}{18}\notag\\
&\Rightarrow\lim_{x \to 0}\frac{x\sin x\tan x - x^{2}\sin x - x^{3} + x^{3} - \sin^{2}x\tan x + x\sin^{2}x}{x^{7}} = \frac{1}{18}\notag\\
&\Rightarrow\lim_{x \to 0}\frac{x\sin x\tan x - x^{2}\sin x + x\sin^{2}x - x^{3}}{x^{7}} - \frac{\sin^{2}x\tan x - x^{3}}{x^{7}} = \frac{1}{18}\notag\\
&\Rightarrow\lim_{x \to 0}\frac{\sin x\tan x - x\sin x + \sin^{2}x - x^{2}}{x^{6}} - L = \frac{1}{18}\notag\\
\end{align}
Our job is done if we can show that $$\lim_{x \to 0}\frac{\sin x\tan x - x\sin x + \sin^{2}x - x^{2}}{x^{6}} = \frac{11}{90}\tag{4}$$ Multiplying $(1)$ and $(2)$ we get $$\lim_{x \to 0}\frac{x\tan x - x^{2} - \sin x\tan x + x\sin x}{x^{6}}= \frac{1}{18}\tag{5}$$ Adding $(4)$ and $(5)$ we see that our proof is complete if we show that $$\lim_{x \to 0}\frac{x\tan x + \sin^{2}x - 2x^{2}}{x^{6}} = \frac{8}{45}\tag{6}$$ Squaring $(1)$ we get $$\lim_{x \to 0}\frac{\sin^{2}x + x^{2} - 2x\sin x}{x^{6}} = \frac{1}{36}\tag{7}$$ Subtracting $(7)$ from $(6)$ we see that proof is complete if we show that $$\lim_{x \to 0}\frac{\tan x + 2\sin x - 3x}{x^{5}}= \frac{3}{20}\tag{8}$$ It is this limit which we will calculate using LHR as follows
\begin{align}
A &= \lim_{x \to 0}\frac{\tan x + 2\sin x - 3x}{x^{5}}\notag\\
&= \lim_{x \to 0}\frac{\sec^{2} x + 2\cos x - 3}{5x^{4}}\text{ (apply LHR)}\notag\\
&= \lim_{x \to 0}\frac{1 + 2\cos^{3} x - 3\cos^{2}x}{5x^{4}\cos^{2}x}\notag\\
&= \frac{1}{5}\lim_{x \to 0}\frac{1 + 2\cos^{3} x - 3\cos^{2}x}{x^{4}}\notag\\
&= \frac{1}{5}\lim_{x \to 0}\frac{(\cos x - 1)^{2}(2\cos x + 1)}{x^{4}}\notag\\
&= \frac{3}{5}\lim_{x \to 0}\left(\frac{1 - \cos x}{x^{2}}\right)^{2}\notag\\
&= \frac{3}{5}\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{3}{20}
\end{align}
Thus the proof is complete by application of LHR three times (once in proof of $(1)$, $(2)$, $(8)$ each). Also note that if you know the result $(1)$ then result $(2)$ can be derived from $(1)$ by subtraction and noting that the limit $$\lim_{x \to 0}\frac{\tan x - \sin x}{x^{3}}$$ can be calculated without LHR very easily. See this question. So in reality we only need two application of LHR for this problem.
Update: While dealing with limit expressions of type $\lim_{x \to 0}f(x)/x^{n}$ for large $n$ (here $n = 7$), I have often found it useful to multiply several well knows limits of type $g(x)/x^{m}$ with smaller values of $m$ to get something like $h(x)/x^{n}$. Expectation is that some terms of $f(x)$ match with those of $h(x)$ and a subtraction would cancel these terms. Also it is expected that resulting expression will be simplified to $p(x)/x^{r}$ when $r < n$. Continue this till we get very small values of exponent of $x$ in denominator. Here for example I have reduced an expression with $x^{7}$ to finally an expression with $x^{5}$ in denominator. See this technique applied to $$\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$$ here. Another application of the same technique can be found here as well.
|
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|
A supposed to be easy calculus problem Find the values of $m$ if the line $y=mx+2$ is a tangent to the curve $x^2-2y^2=1$.
My working:
First we differentiate $x^2-2y^2=1$ with respect to $y$ to get the gradient. We get $y^2=\frac{1}{2}x^2-\frac{1}{2}\implies y=\pm\sqrt{\frac{1}{2}x^2-\frac{1}{2}}$.
We take the positive one for demonstration
$\frac{dy}{dx}=\frac{1}{2}x(\frac{1}{2}x^2-\frac{1}{2})^{-\frac{1}{2}}=\frac{x}{2\sqrt{\frac{1}{2}x^2-\frac{1}{2}}}$
$\implies(1-2m^2)x^2=-2m^2$
Since the tangent touches the curve, we can make $x^2-2(mx+2)^2=1$, we then get $(1-2m^2)x^2=9+8mx$
$\implies(1-2m^2)x^2=-2m^2$ and $(1-2m^2)x^2=9+8mx$ are two equations with two unknowns, then we should be able to find the values of $m$, but I couldn't find any easy way to solve those 2 simultaneous equations. Is there any easier method?
I tried solving $9+8mx=-2m^2$ but we still have two unknowns in one equation?
Also, if we don't use those two simultaneous equations, can we solve this question with a different method?
I am trying to solve WITHOUT implicit differentiation.
Many thanks for the help!
|
One more simplest way:
Put $y=mx+2$ in the equation $x^2-2y^2=1$. Then it comes to a quadratic equation of $x$. From which we get two values of $x$. Since the line is tangent to the given hyperbola so, it can not intersect at two different points. So, the quadratic equation must give two identical values of $x$.
For this, put discriminant is equal to $0$.
Quadratic equation becomes , $x^2-2(mx+2)^2=1$. Putting discriminant equal to $0$ we get, $$64m^2+36(1-2m^2)=0\implies m=\pm \frac{3}{\sqrt 2}$$
|
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|
Probability of $ax^2 + bx + c = 0$ having real solutions
$a$, $b$, $c$ are random integer numbers between $1$ and $100$ (including $1$ and $100$, and uniformly distributed). What is the probability that the equation $ax^2 + bx + c = 0$ has real solutions?
This is from a final high school math exam, and I don't know how to get the answer without using computer.
|
To get a real solution, the discriminant $b^2-4ac\geq0$, so $b\geq 2 \sqrt{ac}$, and $\frac{b^2}{4}\geq ac$. All of our variables are real so there is no need to consider negatives.
Values of $a$ and $b$ can range from $1$ to $n$. The probability of each separate combination is $\frac{1}{n}*\frac{1}{n}$. Considering each combination of $a$ and $b$ separately:
The number of values for $c$ that will give a real solution is $\bigg\lfloor \lfloor \frac{b^2}{4} \rfloor / a \bigg\rfloor$ (restricted to less than 100), so the probability that a random $c$ will give a real solution is $\bigg\lfloor \lfloor \frac{b^2}{4} \rfloor / a \bigg\rfloor * \frac{1}{n}$.
The probability that a random $a$ and random $c$ will give a real solution is $\bigg\lfloor \lfloor \frac{b^2}{4} \rfloor / a \bigg\rfloor * \frac{1}{n} * \frac{1}{n}$.
So, for each possible value of $b$, the probability of a real solution existing is $\frac{1}{n^2}\sum_{a=1}^{n} \bigg\lfloor \lfloor \frac{b^2}{4} \rfloor / a \bigg\rfloor$.
Then, for all possible values of b, the probability of a real solution existing is $\frac{1}{n^3}\sum_{b=1}^{100}\sum_{a=1}^{100} \bigg\lfloor \lfloor \frac{b^2}{4} \rfloor / a \bigg\rfloor$.
I calculated this in Excel for $n=100$ and also got $0.249222$ as did several commenters. I do not see how one would be expected to do this by hand. There is no immediately obvious pattern. Perhaps trying smaller $n$ might offer insight.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $ \int_0^\theta \cosh(a\sin x) dx$ The integral below seems quite simple, but I couldn't find anywhere the result.
$$ I = \int_0^\theta \cosh(a\sin x) dx$$
I tried to expand it into Taylor expansion series and successfully evaluate the integral, but it just got mess,
$$ I =\sum_{k=0}^{\infty} \frac{a^{2k}}{(2k)!} \left[ \frac{1}{2^{2k}}\binom{2k}{k}\theta + \frac{(-1)^k}{2^{2k-1}}\sum_{n=0}^{k-1}(-1)^n\binom{2k}{n} \frac{\sin[(2k-2n)\theta]}{2k-2n}\right]. $$
Is there any simpler form of this integral?
Any helps or hints will be appreciated!
Edited: $\theta$ can only have value of $0 < \theta < \pi/2$.
|
$\int_0^\theta\cosh(a\sin x)~dx=\int_0^\theta\sum\limits_{n=0}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}~dx=\int_0^\theta\left(1+\sum\limits_{n=1}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}\right)~dx$
For $n$ is any natural number,
$\int\sin^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
$\therefore\int_0^\theta\left(1+\sum\limits_{n=1}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}\right)~dx$
$=\left[\sum\limits_{n=0}^\infty\dfrac{a^{2n}x}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{a^{2n}((k-1)!)^2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}\right]_0^\theta$
$=\sum\limits_{n=0}^\infty\dfrac{a^{2n}\theta}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{a^{2n}((k-1)!)^2\sin^{2k-1}\theta\cos\theta}{4^{n-k+1}(n!)^2(2k-1)!}$
$=\theta I_0(a)-\sum\limits_{k=1}^\infty\sum\limits_{n=k}^\infty\dfrac{a^{2n}((k-1)!)^2\sin^{2k-1}\theta\cos\theta}{4^{n-k+1}(n!)^2(2k-1)!}$
$=\theta I_0(a)-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n+2k+2}(k!)^2\sin^{2k+1}\theta\cos\theta}{4^{n+1}((n+k+1)!)^2(2k+1)!}$
Or you can express in terms of Incomplete Bessel Functions:
Consider $$\begin{align}J_0(ia,w)&=\dfrac{2}{\pi}\int_0^w\cos(ia\cos x)~dx
\\&=\dfrac{2}{\pi}\int_0^w\cosh(a\cos x)~dx
\\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cosh\left(a\cos\left(x-\dfrac{\pi}{2}\right)\right)~d\left(x-\dfrac{\pi}{2}\right)
\\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{2}{\pi}\int_0^\frac{\pi}{2}\cosh(a\sin x)~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{2}{\pi}\int_\frac{\pi}{2}^0\cosh\left(a\sin\left(\dfrac{\pi}{2}-x\right)\right)~d\left(\dfrac{\pi}{2}-x\right)
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{2}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx-\dfrac{1}{\pi}\int_\pi^\frac{\pi}{2}\cosh(a\cos(\pi-x))~d(\pi-x)
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx-\dfrac{1}{\pi}\int_\frac{\pi}{2}^\pi\cosh(a\cos x)~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{1}{\pi}\int_0^\pi\cosh(a\cos x)~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-I_0(a)\end{align}
$$
Then $\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx=I_0(a)+I_0(a,w)$
$\therefore\int_0^\theta\cosh(a\sin x)~dx=\dfrac{\pi}{2}\left(I_0(a)+I_0\left(a,\theta-\dfrac{\pi}{2}\right)\right)$
|
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|
Prove that $f(x,y)$ is continuous in $(0,0)$ Prove that $f(x,y)$ is continuous in $(0,0)$, where
\begin{equation}
f(x,y) = \begin{cases}
\frac{x^2y}{x^4+y^2}, & (x,y)\neq 0\\
0, & (x,y) = (0,0)
\end{cases}
\end{equation}
The solution I have is that f is not continuous in $(0,0)$. (The solution doesn't say more than that.)
However, the result I got is that $f$ is continuous in $(0,0)$. Here's my approach:
Lets transform $x$ and $y$ into their polar coordinates, so that we can approach $(0,0)$ from any direction by varying $\theta$:
$(x,y) = (r\cos(\theta),r\sin(\theta)) \qquad r\in\mathbb{R}^+_0 \quad \theta\in[0,2\pi)$
Then $f$ is continuous iff
\begin{equation}
\lim_{(x,y)\to (0,0)} f(x,y) = 0 = f(0,0)
\end{equation}
By using the polar coordinates and letting $r\to 0$ we get:
\begin{equation}
\lim_{(x,y)\to (0,0)} f(x,y)
=
\lim_{r\to 0} \frac{r^3\cos^2\theta\sin\theta}{r^4\cos^4\theta + r^2\sin^2\theta}
= \lim_{r\to 0} \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta}
\end{equation}
By a case distinction by $\theta$ we get:
*
*$\theta\in[0,2\pi)\backslash\{0,\pi\}$:
\begin{equation}
\lim_{r\to 0} \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta}
=
\lim_{r\to 0} \frac{0}{\sin^2\theta} = 0 = f(0,0)
\end{equation}
*$\theta\in\{0,\pi\}$: $\Longrightarrow \sin\theta = 0$
\begin{equation}
\lim_{r\to 0} \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta}
=
\lim_{r\to 0} \frac{0}{r^2\cos^4\theta} = 0 = f(0,0)
\end{equation}
Form 1. and 2. we can conclude that $f$ is continuous in $(0,0)$.
What am I doing wrong in my approach?
|
What you're doing wrong is that you're separating both $r$ and $\theta$ variables. This is not following the continuity definition for functions with several variables.
Please recall the definition of continuity for a function of several variables.
|
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|
Get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$
How to get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$?
*
*I squared the whole denominator, but that didn't help.
*Also I searched for a propriety or identity like $A^2-B^2$, but I didn't see one that could fit.
Any help is appreciated.
|
The trick here is to multiply and divide by the conjugate.
For instance $$\frac{1}{1+\sqrt{2}} = \frac{1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})} = \frac{1-\sqrt{2}}{-1}$$
We can do the same here, but we will do it twice.
$$\frac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}} = \frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{((\sqrt{7}-2\sqrt{5})+\sqrt{3})((\sqrt{7}-2\sqrt{5})-\sqrt{3})} $$
$$=\frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{(\sqrt{7}-2\sqrt{5})^2-3} = \frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{30-4\sqrt{35}}$$
Now multiply and divide by $30+4\sqrt{35}$ and you will elliminate the radical here as well.
|
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|
Decomposition into partial fractions to compute an integral I'm having problems with:
$$\int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}dx$$
I was thinking: $\frac{x^4+1}{x^6+1}$ is an even function and the interval $(-\infty,\infty)$ is symmetric about 0, we could write the integral like:
$$2\int_{0}^{\infty}\frac{(x^4+1)}{x^6+1}dx$$
And for $\frac{x^4+1}{x^6+1}$, I will use the partial fractions. I will write $x^6+1$ like a sum of cubes $x^6+1=(x^2)^3+1^3$ and use the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ and in the end we will have
$$x^6+1=(x^2+1)(x^4-x^2+1)$$ But now I'm stuck and do not know how to decompose $x^4-x^2+1$. I was thinking about $(x^2-1)x^2+1$ or $(x^2-\frac{1}{2})^2+\frac{3}{4}$ but I need a form for that like $-(-1+\sqrt3 x-x^2) (1+\sqrt3 x+x^2)$...
A little help here?
|
Another way,
Solution :
\begin{align}
\int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}\,\mathrm dx&=2\int_{0}^{\infty}\frac{x^4+1}{x^6+1}\,\mathrm dx\\[7pt]
&=2\int_{0}^{\infty}\frac{x^4\,\mathrm dx}{x^6+1}+2\int_{0}^{\infty}\frac{\mathrm dx}{x^6+1}\tag{$\color{red}{❤}$}\\[7pt]
&=\frac{\pi}{3\sin\left(\frac{5\pi}{6}\right)}+\frac{\pi}{3\sin\left(\frac{\pi}{6}\right)}\\[7pt]
&=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{4\pi}{3}}}
\end{align}
Explanation :
$(\color{red}{❤})\;$ $\displaystyle\int_0^\infty\frac{x^{\large n-1}}{x^m+1}\ \,\mathrm dx=\frac{\pi}{m\sin\left(\frac{n\pi}{m}\right)}\,$ for $\,0<n<m$.
|
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|
How to find $\frac{0}{0}$ limit without L'Hôpital's rule I am having trouble solving this limit. I tried applying L'Hôpital's rule but I got $\frac{0}{0}$.
$$\lim_{x\to0} {\frac{\frac{1}{1+x^3} + \frac{1}{3}\log{\left(1+3x^3\right)}-1}{2\sin{\left(3x^2\right)}-3\arctan{\left(2x^2\right)}}}$$
I would appreciate any hints in the right direction. Thanks in advance for your help.
|
Hint: $$\frac{1}{1+x^3} = 1 - x^{3} + x^{6} - x^{9} + O(x^{12}) \,\,\, \text{and} \,\,\, \log (1 + 3x^3) = 3x^3 - \frac{9x^6}{2} - 9x^9 + O(x^{12})$$
$$\sin (3x^2) = 3x^2 - \frac{9x^6}{2} + O(x^{10}) \,\,\, \text{and} \,\,\, \arctan (2x^2) = 2x^2 - \frac{8x^6}{3} + O(x^{10}) $$
Can you take it from here?
|
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|
Find value of $xy\sqrt{y^2 - x^2}$ for the given differential equation.
If $(y^3 - 2x^2y)dx + (2xy^2 - x^3)dy = 0 $ , then prove that the value of $xy\sqrt{y^2 - x^2}$ is a constant.
This is what I've tried :
$$ y(y^2 - 2x^2) dx + x(2y^2 - x^2) dy = 0 \\
\cfrac{dy}{dx} = \cfrac{(2x^2 - y^2)y}{(2y^2 - x^2)x} \\
$$
I then tried substituting $y/x = v$
Further, solving it, I got :
$$xy\sqrt{y^2 - x^2} = \cfrac{x^{4/3}}{y^{4/3}}$$
But, I'm not sure how to come with a constant at RHS.
|
When you substitute the change of variables you obtain:
$$ xu' = u \left( \frac{2-u^2}{2u^2-1} -1 \right), $$
which upon integration and taking exponential in both sides gives (I used Mathematica):
$$ x e^C = \frac{1}{u^{1/3} ({1-u^2})^{1/6}}, $$ where $e^C :=K$ is a constant of integration. Substitute back $u = y/x$ to find:
$$ K x = \frac{1}{\left(\frac{y}{x} \right)^{1/3} \left({1-\frac{y^2}{x^2}} \right)^{1/6}},$$ now cube everything to come up with:
$$ K x^2 y \left( 1-y^2/x^2 \right)^{1/2} = K x y \sqrt{x^2-y^2 }=1, $$ which is equivalent to the desired result.
Hope this helps!
|
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|
Difficult inverse tangent identity
Prove that:
$$\arctan\left(\frac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}} \right) = \frac{\pi}{4} - \frac{1}{2}\arccos(x), -\frac{1}{\sqrt{2}} \le x \le 1$$
I'd multiply the inside of $\arctan$ by the conjugate of the denominator.
I get:
$$\arctan\left(\frac{1 - 1\sqrt{1 - x^2}}{x} \right)$$
But that is still very difficult.
Any HINTS, no solutions?
|
In the other way round, let
$$\arctan\left(\frac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}} \right) =y$$
$\implies -\dfrac\pi2\le y\le\dfrac\pi2\ \ \ \ (1)$ and $\tan y=\dfrac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}}$
Applying Componendo and dividendo, $$\dfrac{\sqrt{1 + x}}{\sqrt{1 - x}}=\dfrac{1+\tan y}{1-\tan y}$$
Squaring we get, $$\dfrac{1+x}{1-x}=\dfrac{1+\tan^2y+2\tan y}{1+\tan^2y-2\tan y}$$
Again applying Componendo and dividendo, $$x=\dfrac{2\tan y}{1+\tan^2y}=\sin2y$$
Now $-\dfrac1{\sqrt2}\le x\le1\implies-\dfrac\pi4\le2y\le\dfrac\pi2\ \ \ \ (2)$ (using $(1)$)
Finally, $2y=\arcsin x$ if $-\dfrac\pi2\le2y\le\dfrac\pi2$ which is satisfied by $(2)$
$\implies2y=\arcsin x=\dfrac\pi2-\arccos x$
|
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|
How to show algebraically that $x^3 +3x +1$ is injective?
How to show algebraically that $$x^3 +3x +1$$ is injective?
Working with the usual method of assuming that $f(c)=f(d)$ and then seeing if $c=d$. I've tried several approaches, including factoring by the difference of cubes, followed by use of the quadratic formula. I'm stumped. Any help greatly appreciated!
Thanks, Bob
|
Let $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^3+3x+1$, let's suppose $x$ and $y$ are real numbers, then
\begin{align*}
f(x)=f(y)\quad \iff\quad x^3+3x+1&=y^3+3y+1\\
x^3-y^3+3(x-y)&=0
\end{align*}
We can factor $x^3-y^3$ as $(x-y)(x^2+xy+y^2)$, so
\begin{align*}
f(x)=f(y)\quad \iff(x-y)(x^2+xy+y^2+3)&=0
\end{align*}
Since $$x^2+xy+y^2=(x+\frac{1}{2}y)^2+\frac{3}{4}y^2\ge 0$$ we have $$x^2+xy+y^2+3\ge 3$$
Then
\begin{align*}
f(x)=f(y)\quad \iff x-y&=0 \iff x=y
\end{align*}
|
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|
How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$? let $m,n$ be integers, show that if $ n>m\geq 0 $ :
$$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3}
{2}\left(\frac{1}{\sqrt{3}}\right)^{n-m}$$
where real $x,y,z > 0 $ and $xy + yz + zx = 1$
Thank you for your help .
|
Despite the attempts in the other answers and comments, and the two ``proofs'' given there, the inequality does not hold in general. We give two proofs for its failure:
First Proof (involving huge numbers): Set $x=3/7$, $y=4/7$, $z=37/49$. Then $xy+yz+zx=1$. For $m=7$, $n=8$ set $A=\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}$ and $B=\frac{3}{2}(\frac{1}{\sqrt{3}})^{n-m}$. Then
\begin{equation}
A^2-B^2=-\frac{5464419604082977128654242570410694589510448147711713}{929486260504473222256638487813651283882185173994428050}<0.
\end{equation}
Second proof (more conceptual): Suppose that $0<x<y<z$. Then, with $n=m+1$,
\begin{equation}
\lim_{m\to\infty}
(\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m})
=
\lim_{m\to\infty}
(\frac{x}{1+(y/x)^m}+\frac{y}{1+(z/y)^m}+\frac{z}{1+(x/z)^m})=z
\end{equation}
Now set $x=\frac{1}{3}$, $y=\frac{2}{3}$, $z=\frac{7}{9}$. Then $xy+yz+zx=1$, but $z<\frac{3}{2}\frac{1}{\sqrt{3}}$. So the stated inequality does not hold for all sufficiently large $m$.
Remark: The same question was asked here at MathOverflow, and I gave a similar counterexample there.
|
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|
Solving $\cos x+\sin x-1=0$ How does one solve this equation?
$$\cos {x}+\sin {x}-1=0$$
I have no idea how to start it.
Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?
Thanks in advance!
|
Given $$\color{blue}{\cos x+\sin x-1=0} $$$$\cos x+\sin x=1 $$ Divide both sides by $\color{blue}{\sqrt{2}}$ we get $$\frac{1}{\sqrt{2}}\cos x+ \frac{1}{\sqrt{2}}\sin x=\frac{1}{\sqrt{2}}$$ $$\cos x\cos\frac{\pi}{4}+\sin x\sin\frac{\pi}{4}=\cos\frac{\pi}{4}$$ Using formula $\color{purple}{\cos A\cos B+\sin A\sin B=\cos(A-B)}$, we get $$\color{green}{\cos\left(x-\frac{\pi}{4}\right)=\cos\frac{\pi}{4}}$$ As there is no information about the unknown value $x$ hence writing the general solutions as follows $$x-\frac{\pi}{4}=2n\pi\pm \frac{\pi}{4}$$$$x=2n\pi\pm \frac{\pi}{4}+\frac{\pi}{4}$$$$ \color{}{x=2n\pi} \quad \text{Or}\quad \color{}{x=2n\pi+\frac{\pi}{2}} $$
$$\color{blue}{x\in\{2n\pi\}\cup\{2n\pi+\frac{\pi}{2}\}}$$ Where, $\color{}{n \space \text{is any integer}}$ i.e. $\ n=0, \pm1, \pm2,\pm3, \ldots$
|
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|
Why cannot $(2x^2 + x)^2$ be simplified to $2x^4 + x^2$? So, I want to simplify an equation : $(2x^2 + x)^2$.
I thought this would simplify to $2x^4 + x^2$
But, if you input a value for $x$, the answers do not equal. For example, if you input $x = 3$, then:
$$(2x^2+x)^2
= 21^2
= 441$$
AND:
$$2x^4 + x^2
= 2(82) + 9
= 173$$
Can anyone explain why this is the case?
|
Think of it like this.
$a^{2}$ means $a \times a$.
More generally, $a^{n}$ means $\underbrace{a \times a \times a \times \dots \times a}_{n \text{ times}}$.
So $2^{3}$ means $2 \times 2 \times 2$. That's $2 \times 2$, $3$ times.
Therefore $(2x^{2} + x)^{2}$ means $(2x^{2} + x) \times (2x^{2} + x)$.
Now, using your number, we see that:
$(2 \times 3^{2}) + 3 = 18 + 3 = 21$
And $21 \times 21$ is very different to $(2 \times 3^{4}) + 3^{2} = 162 + 9 = 171$.
What you have learned the hard way is that exponentiation is not distributive over addition. Which means that $(a + b)^{2} \neq a^{2} + b^{2}$. Why is this? Because $(a + b)^{2}$ means:
Multiply $(a + b)$ by $(a + b)$
Not multiply $a$ by $a$ and multiply $b$ by $b$ and add them together.
For a more simple example than yours, let's look at $(3 + 2)^{2}$
$(3 + 2)^{2} = (3 + 2) \times (3 + 2) = 5 \times 5 = 25$
But:
$(3 + 2)^{2} \neq 3 \times 3 + 2 \times 2 = 9 + 4 = 13$
N.B.: Here we are talking about the general case. There is certainly one specific example of where $(a + b)^{2} = a^{2} + b^{2}$ but this is basically just by coincidence.
|
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|
What is wrong with this integral reasoning? $$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx$$
We start by multiplying by $1=\frac{x}{x}$.
$$\int\frac{x^{2}+1}{x^{2}\sqrt{x^{4}+1}}xdx$$
Next, we use the substitution $u=x^{2}$;$\frac{du}{2}=xdx$.
$$\frac{1}{2}\int\frac{u+1}{u\sqrt{u^{2}+1}}du=\frac{1}{2}\int\frac{1}{\sqrt{u^{2}+1}}du+\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du=$$
$$=\frac{1}{2}\ln\left(u+\sqrt{u^{2}+1}\right)+\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du$$
The problem is now reduced to computing the integral $\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du$.
$$\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du=\frac{1}{2}\int\frac{1}{u^2\sqrt{1+\frac{1}{u^{2}}}}du$$
We use substitution again $v=\frac{1}{u}$;$-dv=\frac{1}{u^{2}}du$, then we have the next integral.
$$-\frac{1}{2}\int\frac{1}{\sqrt{1+v^{2}}}dv=-\frac{1}{2}\ln\left(v+\sqrt{1+v^{2}}\right)=-\frac{1}{2}\ln\left(\frac{1}{u}+\sqrt{1+\frac{1}{u^{2}}}\right)=$$
$$=-\frac{1}{2}\ln\left(\frac{1+\sqrt{u^{2}+1}}{u}\right)$$
Finally the solution is:
$$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx=\frac{1}{2}\ln\left(u+\sqrt{1+u^{2}}\right)-\frac{1}{2}\ln\left(\frac{1+\sqrt{u^{2}+1}}{u}\right)=\frac{1}{2}\ln\left(\frac{x^{4}+x^{2}\sqrt{x^{4}+1}}{1+\sqrt{x^{4}+1}}\right).$$
But the problem is, the book has the following solution:
$$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx=\ln\left(\frac{x^{2}-1+\sqrt{x^{4}+1}}{x}\right)$$
which is obviously different.
|
I have differentiated your answer and the book's answer. They are both equal to the integrand. So your work is indeed correct.
In fact, the book's answer can be considered wrong since it lacks an absolute value around the argument to $ln$, whereas your answer does not need one there since the argument to $ln$ in your expression is strictly positive.
|
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|
$(1+i)^6$ in polar form $re^{i\theta}$ I used De Moivre's formula and got
$$
\left(\frac{\sqrt{2}}{2}\right)^6 \times
\cos\left(6 \times \frac{1}{4\pi}\right) +
i\sin\left(6 \times \frac{1}{4\pi}\right) =
\frac{1}{8} e^{\frac{3}{2\pi}}.
$$
But the answer is $8e^{\frac{3}{2\pi}}$, can someone explain where the $8$ is from?
Thanks!
|
$$
\left(\frac{\sqrt 2}2\right)^6 = \left(\frac{\sqrt 2}{\sqrt 2\sqrt 2}\right)^6 = \left( \frac 1 {\sqrt 2} \right)^6 = \frac 1 {\sqrt{2}^6} =\frac 1 {(\sqrt 2^2)^3} = \frac 1 {2^3} = \frac 1 8.
$$
(But you have $\dfrac 1{4\pi}$ where you need $\dfrac 1 4 \pi$.)
|
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|
Generating Functions Dice In how many ways can we roll a red die, a yellow die, and a black die, and get a sum of $9$?
(The dice are colored so that a red $2$, a yellow $3$, and a black $4$ is distinguished from a red $3$, a yellow $4$, and a black $2$, for example.)
Should I use generating functions somehow?
|
Yes. Multiply out:
$$(x+x^2+x^3+x^4+x^5+x^6)^3$$
and find the coefficient of $x^9$. Do you see why this works?
EDIT: I'll elaborate more for the sake of completeness. It is because when you multiply:
$$\color{red}{(x+x^2+x^3+x^4+x^5+x^6)}\color{yellow}{(x+x^2+x^3+x^4+x^5+x^6)}(x+x^2+x^3+x^4+x^5+x^6)$$
Sorry for the yellow color, but treat each of these factors as one of the dice. Now to get a sum of $9$ on the dice, you are effectively picking powers of $x$ from each of the factors such that their exponents sum to $9$. The number of times you can pick powers of $x$ from each of the factors like that (which is the number of ways we can get a sum of $9$ on the dice) is exactly the coefficient of $x^9$ when the product is multiplied out.
|
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|
Show $\int \dfrac{\sinh x}{\cosh 2x}=\dfrac 1 {2\sqrt 2} \ln\left|\dfrac {\sqrt 2 \cosh x-1}{\sqrt 2 \cosh x +1}\right| + C$
Show by means of the substitution $u = \cosh x$, that
$$\int \dfrac{\sinh x}{\cosh 2x}=\dfrac 1 {2\sqrt 2} \ln\left|\dfrac {\sqrt 2 \cosh x-1}{\sqrt 2 \cosh x +1}\right| + C$$
$$\int \dfrac{\sinh x}{\cosh 2x}= \dfrac {\sqrt 2}{2} \tanh ^{-1} (\sqrt 2) u + C$$
Let $$\tanh ^{-1} (\sqrt 2u) = y$$
Then $$2y = \ln \left| - \dfrac {\sqrt 2 u +1}{\sqrt 2 u -1}\right|$$
Given $\sqrt 2 u -1 \geq 0$ or since $u = \cosh x \geq 1$
$$y = \dfrac 1 2 \ln \left|\dfrac {\sqrt 2 u +1}{\sqrt 2 u -1}\right|$$
Giving
$$\int \dfrac{\sinh x}{\cosh 2x}=\dfrac 1 {2\sqrt 2} \ln\left|\dfrac {\sqrt 2 \cosh x+1}{\sqrt 2 \cosh x -1}\right| + C$$
Which is different to the given solution.
Where did I go wrong?
Many thanks in advance
|
I believe your mistake is missing the negative sign here:
$$\int \dfrac{\sinh x}{\cosh 2x}= -\dfrac {\sqrt 2}{2} \tanh ^{-1} (\sqrt 2) u + C$$
In lots more steps:
$$\begin{split}
\int \dfrac{\sinh x}{\cosh 2x}dx &= \int \dfrac{\sinh x}{2\cosh^2(x) - 1}dx \\
&= \int \dfrac{1}{2u^2 - 1}du \\
&= \dfrac{1}{\sqrt2}\int \dfrac{1}{v^2 - 1}dv \\
&= -\dfrac{1}{\sqrt2}\int \dfrac{1}{1 - v^2}dv (*) \\
&= -\dfrac{1}{\sqrt2}\tanh^{-1} v + C \\
&= -\dfrac{1}{\sqrt2}\tanh^{-1}(u\sqrt{2}) + C
\end{split}$$
Need to invert in $(*)$ because $\dfrac{d}{dx}\tanh^{-1}x = \dfrac{1}{1 - x^2}$. The rest follows from the missing negative.
|
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|
Coefficient Problem (polynomial expansion)
Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$
Just to begin,
$(1-x)(1+2x) = -2x^2 + x + 1$
$(1-x)(1+2x)(1-3x) = 6x^3 - 5x^2 - 2x + 1$
But expanding on like this take too long.
In the end the terms will be like:
$(1 - 15x)(a_0 + a_1x^1 + a_2x^2 + ....)$
Then just looking at the $x^2$, it will be:
$(a_2x^2 - 15a_1x^2) = x^2(a_2 - 15a_1)$
But it it still a very hard problem, any hints?
|
Another approach: consider the product
$$(1-x)(1+2x)\cdots(1+(-1)^nnx)=1+a_nx+b_nx^2+\cdots\ .$$
It is easy to see that the coefficient of $x$ in this expression is
$$\eqalign{a_n
&=-1+2-3+4-\cdots+(-1)^nn\cr
&=\cases{(-1+2)+(-3+4)+\cdots+(-(n-1)+n)&if $n$ is even\cr
-1+(2-3)+(4-5)+\cdots+((n-1)-n) &if $n$ is odd\cr}\cr
&=\cases{k&if $n=2k$\cr -1-k&if $n=2k+1$\cr}\cr
&=(-1)^n\left\lceil\frac n2\right\rceil\ .\cr}$$
Hence the coefficient of $x^2$ satisfies the recurrence
$$b_{n+1}=b_n+(-1)^{n+1}(n+1)a_n\ ,$$
that is,
$$b_{n+1}=b_n-(n+1)\left\lceil\frac n2\right\rceil\ .$$
It is now pretty easy to calculate by mental arithmetic the values
$$\eqalign{
b_1,b_2,\ldots,b_{15}&=0,-2,-5,-13,-23,-41,-62,\cr
&\qquad{}-94,-130,-180,-235,-307,-385,-483,-588.\cr}$$
|
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|
Prove that there are no positive integers $a$ and $b$, such that $b^4 + b + 1 = a^4$ So I've been trying to solve this problem for a couple of days now. What I've come up with is this:
By way of contradiction, assume that there are positive integers a and b such that $b^4 + b + 1 = a^4$.
Consider the case when $b \geq a$. Then $a^4 - b^4 = b + 1$, but also $a^4 - b^4 \leq 0$. Thus, $b + 1 \leq 0$. Then, $b \leq -1$, a contradiction.
However, I haven't been able to figure out how to prove it with the case of $b < a$.
Thank you.
|
In any case, $b^4-a^4+b+1=0$. Also,
$$
(b^4-a^4)=(b^2-a^2)(b^2+a^2)=(b-a)(b+a)(b^2+a^2)
$$
So,
$$
(b-a)(b+a)(b^2+a^2)= -b-1
$$
So $(b^2+a^2)$ now divides $b+1$. So $b^2+a^2\leq b+1$. The only way this will work is if $a=b=1$. Check that this doesn't satisfy the original equation. So we're done!
|
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|
If for a prime p $1+\frac{1}{2}+\frac{1}{3}+\ldots + \frac{1}{p-1}=\frac{a}{b}$ then show that p divides a. Moreover if $p>3$ then $p^2$ divides a. Here is a problem from Hersteins Topics in Algebra:
If $1+\frac{1}{2}+\frac{1}{3}+\ldots + \frac{1}{p-1}=\frac{a}{b}$ for a prime $p$, then show that $p$ divides $a$. Moreover if $p>3$ then $p^2$ divides $a$.
I am able to prove the first part by writing
$$1+\frac{1}{2}+\frac{1}{3}+\ldots + \frac{1}{p-1}=\frac{(p-1)!+ \frac{(p-1)!}{2}+\frac{(p-1)!}{3}+\ldots + 1 }{(p-1)!},$$
where we can show the numerator to be congruent to $0$ mod $p$. I can't seem to crack the second part. Thanks in advance for any help.
|
The numerator is $(p-1)!\left ( \frac 11 + \frac 12 + ... + \frac 1{p-1}\right ) \equiv (p-1)!\left ( 1^{-1} + 2^{-1} + ... + (p-1)^{-1}\right ) \mod {p^2}$.
The Euler function for $p^2$ is $p(p-1)$. Thus, for any $k \lt p $ , $k^{-1} \equiv k^{p(p-1)-1}\mod p^2$.
Thus, $k^{-1} + (p-k)^{-1} \equiv k^{p(p-1)-1} + (p-k)^{p(p-1)-1} \equiv k^{p(p-1)-1} + p^2.c + (-k)^{p(p-1)-1} $ ($\because p \gt 3$). Also, as $p(p-1)-1$ is odd, hence $(-k)^{p(p-1)-1} \equiv (-1).(k)^{p(p-1)-1}$. Thus, $k^{p(p-1)-1} + (p-k)^{p(p-1)-1} \equiv 0 \mod p^2$. Hence numerator is divisible by $p^2$.
|
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|
Geometry Problem Isosceles Triangle
Given this isosceles triangle, find angle AMC.
|
By using Ceva's theorem
$$\frac{AE}{EB} \cdot \frac{BQ}{QC} \cdot \frac{CP}{PA} = 1$$
$$\frac{\sin\angle ACE}{\sin\angle ECB}\cdot \frac{\sin 10^\circ}{\sin 40^\circ}
\cdot \frac{\sin 20^\circ}{\sin 30^\circ}=1$$
$$\frac{\sin(80^\circ-\angle ACE)}{\sin\angle ACE}=\frac{\sin 10^\circ}{\sin 40^\circ}\cdot \frac{\sin 20^\circ}{\sin 30^\circ}=
\frac{\sin 10^\circ}{2\sin 20^\circ\cos 20^\circ}\cdot \frac{\sin 20^\circ}{\frac{1}{2}}=
\frac{\sin 10^\circ}{\cos 20^\circ}$$
$$\sin 80^\circ\cot\angle ACE - \cos 80^\circ = \frac{\sin 10^\circ}{\cos 20^\circ}$$
Hence we find $\angle ACE =70^\circ$ and $\angle AMC = 180^\circ - 70^\circ - 40^\circ = 70^\circ$
|
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|
Difficult Integration $\frac{1}{(2x-1)^{\frac {1}{2}}-(2x-1)^{\frac{1}{4}}}$ I am trying to solve this integral:$$\int{\frac{1}{(2x-1)^{\frac {1}{2}}-(2x-1)^{\frac{1}{4}}}}dx$$
My idea was to replace the second term in the denominator as $u$ and therefore have $\int{\frac{1}{u^2-u}}dx$ but it looks still complicated. On WolframAlpha I have a long way to solve it. Do you have any advice of a shorter approach?
|
Since I realized that I think I'm nearly done with a solution, here's the continuation:
For the record, with that $u$-substitution, the integral becomes:
$$u=(2x-1)^\frac{1}{4} \rightarrow x=\frac{u^4+1}{2}$$
$$du=\frac{1}{4}(2x-1)^\frac{-3}{4}2dx$$
$$\int \frac{1}{u^2-u}dx$$
$$\int \frac{1}{u^2-u}\left( 2(2x-1)^\frac{3}{4} \right)du$$
$$\int \frac{1}{u^2-u}\left( 2(u^4)^\frac{3}{4} \right)du$$
$$\int \frac{1}{u^2-u}\left( 2u^3 \right)du$$
$$\int \frac{2u^3}{u^2-u}du$$
$$\int \frac{2u^2}{u-1}du$$
Do polynomial division:
$$\int \left ( 2u+\frac{2}{u-1}+2 \right) du$$
This is trivial to integrate:
$$u^2+2\ln (|u-1|) +2u+C$$
Substitute back to the original $x$ variable:
$$(2x-1)^\frac{1}{2}+2\ln (|(2x-1)^\frac{1}{4}-1|) +2(2x-1)^\frac{1}{4}+C$$
|
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|
Equation of a ... 3D object??? (Stupid question...)
Well we can represent a point as something like $P(a,b,c)$
We can represent a line as $\dfrac{x-a}{p}=\dfrac{x-b}{q}=\dfrac{x-c}{r}$
We can also represent a plane as $ax+by+cz=d$
So can we represent a $3D$ object also???
Or do we need to have have a 4D cartesian system or something like that?
EDIT
I thought this .gif is interesting!
|
We can represent 3D objects as a function of three variables. Here are some simple, finite geometric shapes, defined implicitly in Cartesian coordinates. The coefficients are used to scale the shape, and can be set to any value. You will end up with tall or flat prisms and pyramids by adjusting them away from their current values.
Sphere : $x^2+y^2+z^2 = a^2 $
Torus : $\big(\sqrt{x^2+y^2} - a\big)^2 + z^2 = b^2 $
Triangle Torus : $\left|\big|\big(\sqrt{x^2+y^2} - a\big)^2\big| + 2z\right| + \left|\big(\sqrt{x^2+y^2} - a\big)^2\right| = b$
Square Torus : $\left|\big(\sqrt{x^2+y^2} - a\big)^2-z\right| + \left|\big(\sqrt{x^2+y^2} - a\big)^2+z\right| = b$
Cone : $\left |\sqrt{x^2+y^2} +2z\right | + \sqrt{x^2+y^2} = a$
Cylinder : $\left |\sqrt{x^2+y^2} - z\right | + \left |\sqrt{x^2+y^2} + z\right | = a$
Tetrahedron : $\left|\big||x|+2y\big|+|x| + 2z\right| + \big||x|+2y\big|+|x| = a$
Triangle Prism : $\big |\big||x|+2y\big|+|x| - 2z\big | + \big |\big||x|+2y\big|+|x| + 2z\big | = a $
Square Pyramid : $\big ||x-y|+|x+y| + 3z\big | + |x-y|+|x+y| = a$
Cube : $\big ||x-y|+|x+y| - 2z\big | +\big ||x-y|+|x+y| + 2z\big | = a$
And, of course, these can easily be extended to 4D shapes as well (or any number of dimensions), that are just as elementary.
|
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|
Integral using contour integration Here is the integral I want to evaluate:
$$\int_{0}^{2\pi} \frac{dx}{a+b \cos x }, \quad a>b >0$$
Apparently there are limitations as to what values $a, b$ are supposed to take but let us not concern about this. Since using the sub $u =\tan \frac{x}{2}$ (the Weiersstrass sub) results that the integral is $0$ (as it should, since it is not $1-1$ function in this interval) I got down down the way of contour integration. Hence:
$$\begin{aligned}
\int_{0}^{2\pi}\frac{dx}{a+ b\cos x} &\overset{x=i \ln u}{=\! =\! =\!} \oint \limits_{|z|=1} \frac{dz}{iz \left [ a + \frac{b}{2}\left ( z+z^{-1} \right ) \right ]} \\
&= \frac{1}{i} \oint \limits_{|z|=1} \frac{dz}{za + \frac{bz^2}{2}+\frac{b}{2}}\\
&=\frac{2}{i} \oint \limits_{|z|=1} \frac{dz}{bz^2 +b +2za} \\
&= \frac{2}{i} 2\pi i \sum_\text{residues} f(z)\\
&= \frac{4\pi}{1+ \sqrt{1-8ab}+2a}
\end{aligned}$$
However, judging by intuition this must not be the result. Because this one restricts the integral too much. What i mean is, that for $a=6, b=3$ we have that:
$$\int_0^{2\pi} \frac{dx}{6+3\cos x}= \frac{2\pi}{3\sqrt{3}}$$
My formula cannot derive the result because then radical would be negative. What am I doing wrong here?
|
Your transformation to a contour integral is correct; we note that
$$
\int_0^{2\pi}\frac{dx}{a+b\cos x}=\int_0^{2\pi}\frac{dx}{a+\frac{b}{2}(e^{ix}+e^{-ix})}
$$
and set $z=e^{ix}$ (a parametrization of the unit circle) so that $dz=ie^{ix}dx=izdx\to dx=\frac{dz}{iz}$. This yields
$$
\int_0^{2\pi}\frac{dx}{a+b\cos x}=\oint_{|z|=1} \frac{dz}{iz(a+\frac{b}{2}(z+\frac{1}{z}))}.
$$
The problem is in your residue calculation. By the quadratic formula, the zeros of $bz^2+2az+b$ are $\frac{-a\pm\sqrt{a^2-b^2}}{b}$. By Vieta's formula, the product of the roots is $1$ so only one is within the unit disc; clearly it must be $w=\frac{-a+\sqrt{a^2-b^2}}{b}$. You can then calculate the residue
$$
\text{res}_{w}\left(\frac{1}{bz^2+2az+b}\right) =
\lim_{z\to w}\frac{1}{b\left(z-\frac{-a-\sqrt{a^2-b^2}}{b}\right)}=
\frac{1}{2\sqrt{a^2-b^2}}
$$
This will give you a final result of $\frac{2\pi}{\sqrt{a^2-b^2}}$.
|
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|
inequalities with fraction problem x $\frac{1}{x} > \frac{2x} {x^2 +2}$ solving this inequalities:
My long solution (wrong) :
multiplying $(x^2 + 2)^2 (x)^2 \dots$ (multiplying square of each denominator, getting rid of the > or < 0)
$x (x^2 +2)^2 > 2x(x^2) ( x^2 +2)$
$x(x^4+4x^2 +4)>2x^2(x^2+2)$
$x^5+4x^3+4x>2x^4+4x^2$
$x^5+4x^3+4x-2x^4-4x^2>0$
$x ( x^4 + 4x^2 + 4 +2x^3 -4x) >0 $
$x>0$ or ^ ...
This solving method doesn't look right
|
Since $x^2+2>0$,
$$\frac{1}{x}>\frac{2x}{x^2+2}\iff \frac{x^2+2}{x}=x+\frac{2}{x}>2x\iff x<\frac{2}{x}$$
$$\iff \begin{cases}\begin{cases}x>0\\ x^2<2\end{cases}\\\ \ \ \text{or}\\ \begin{cases}x<0\\ x^2>2\end{cases}\end{cases}\iff \begin{cases}\begin{cases}x>0\\ -\sqrt{2}<x<\sqrt{2}\end{cases}\\\ \ \ \text{or}\\ \begin{cases}x<0\\ x\in(-\infty,-\sqrt{2})\cup(\sqrt{2},+\infty)\end{cases}\end{cases}$$
$$\iff \begin{cases}0<x<\sqrt{2}\\\ \ \ \text{or}\\ x\in(-\infty,-\sqrt{2})\end{cases}$$
|
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The sum $\sum_{k=0}^\infty (k-1)/2^k$ is zero I am trying to prove an equation given in the CLRS exercise book (AP GP clrs appendix A.1-4). The equation is:
$$\sum_{k=0}^\infty (k-1)/2^k=0$$
I solved the LHS but my answer is 1 whereas the RHS should be 0 Following is my solution:
Let's say S = k/2^k = 1/2 + 2/2^2 + 3/2^3 + 4/2^4 ....
2S = 1 + 2/2 + 3/2^2 + 4/2^3 ...
2S - S = 1 + ( 2/2 - 1/2) + (3/2^2 - 2/2^2) + (4/2^3 - 3/2^3)..
S = 1+ 1/2 + 1/2^2 + 1/2^3 + 1/2^4..
S = 2 -- eq 1
Now let's say S1 = (k-1)/2^k = 0/2 + 1/2^2 + 2/2^3 + 3/2^4...
S - S1 = 1/2 + (2/2^2 - 1/2^2) + (3/2^3 - 2/2^3) + (4/2^4 - 3/2^4)....
S - S1 = 1/2 + 1/2^2 + 1/2^3 + 1/2^4...
= 1
From eq 1
2 - S1 = 1
S1 = 1
Whereas the required RHS is 0. Is there anything wrong with my solution? Thanks..
|
You forgot to add the term $-1$ in $S_1$. It is achieved at $k=0$. So, $ S-S_1 = 2.$
|
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|
Is there any general method for solving $(a_1+a_2+..a_n)^2=a_1^3+a_2^3+...+a_n^3$ in positive integers $a_1,a_2,...a_n$? We know the identity $(1+2+...+n)^2=1^3+2^3+...+n^3$ . So I was thinking , for given $n\in \mathbb N$ , is there any general method for solving $(a_1+a_2+..a_n)^2=a_1^3+a_2^3+...+a_n^3$ in positive integers ? At least can we find all positive integers $a,b,c$ such that $(a+b+c)^2=a^3+b^3+c^3$ ?
|
If $(a_1+a_2+\cdots+a_n)^2=a_1^3+a_2^3+\cdots+a_n^3$ and $m=\max a_k$, then $m\le n^2$.
Indeed, $m^3 \le \sum_{k=1}^n a_k^3 = \left( \sum_{k=1}^n a_k \right)^2 \le (nm)^2 = n^2m^2$.
This makes your last question easy to answer:
Indeed, if $(a+b+c)^2=a^3+b^3+c^3$ then $a,b,c \le 3^2=9$.
Testing all possibilities gives exactly two solutions: $(1,2,3)$ and $(3,3,3)$.
|
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|
Finding minimal polynomial of big blocks diagonal matrix Consider the following matrix:
\begin{bmatrix}
-3 & 1 & -1 & & & & & \\
-7 & 5 & -1 & & 0 & & & 0\\
-6 & 6 & -2 & & & & & \\
& & & 4 & 0 & 1 & & \\
& 0 & & 0 & 1 & 0 & & 0\\
& & & 0 & 0 & 4 & & \\
& & & & & & -1 & -1\\
& 0 & & & 0 & & 1 & -3
\end{bmatrix}
I need to find its minimal polynomial.
I found that the characteristic polynomial is: $P(\lambda)=(\lambda+2)^4(\lambda-4)^3(1-\lambda)$ using the fact that for a block matrix:
$$A=\begin{bmatrix} {A}_{1} & 0 & \cdots & 0 \\ 0 & {A}_{2} & \cdots & 0 \\\vdots & \vdots & \ddots & \vdots \\0 & 0 & \cdots & {A}_{n} \end{bmatrix} $$
It holds that: $\det(A)=\prod_{i=1}^{n}\det(A_i)$
Is there an easy trick to find the minimal polynomial as well?
|
$p(A)=0$ is equivalent to $p(A_1)=\dots=p(A_n)=0$
So it is the minimum common multiple of the blocks minimum polynomials.
|
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|
Are the fields $\mathbb{Q}(\sqrt[7]{16}+3 \sqrt[7]{8})$ and $\mathbb{Q}(\sqrt[7]{16})$ equal? I have trouble with these field extensions. Is field $\mathbb{Q}(\sqrt[7]{16}+3 \sqrt[7]{8})$ equal to field $\mathbb{Q}(\sqrt[7]{16})$?
We can $\sqrt[7]{16}+3 \sqrt[7]{8}$ express as $(\sqrt[7]{2}+3)(\sqrt[7]2)^3$.
|
$$(\sqrt[7]{16})^6=8\sqrt[7]8$$
This shows that $\Bbb Q[\sqrt[7]{16}]\supseteq\Bbb Q[\sqrt[7]{16}+3\sqrt[7]8]$.
Now, consider the polynomials $x^4+3x^3$ and $9x^6+2x+6$. They are coprime, so we can apply the Bezout's identity (note that $\Bbb Q[x]$ is an Euclidean domain) to guarantee that there exist some polynomials $P$ and $Q$ such that
$$P(x)(x^4+2x^3)+Q(x)(9x^6+2x+6)=x^4$$
Let $a=\sqrt[7]2$. Then we find that
$$(\sqrt[7]{16}+3\sqrt[7]8)^2=(a^4+3a^3)^2=9a^6+2a+6$$
Therefore,
$$P(a)(a^4+3a^3)+Q(a)(a^4+3a^3)^2=a^4$$
This shows that $\Bbb Q[\sqrt[7]{16}]\subseteq\Bbb Q[\sqrt[7]{16}+3\sqrt[7]8]$
|
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|
Find series power of $F(x) =e^{-x}x^{2}$ i need help for this problem; find a power series for $F(x) \text{=}e^{-x}x^{2} $ and derivate and prove this expression $$ \sum \limits^{\infty }_{n=1}\frac{(-2)^{n+1}(n+2)}{n!} =\text{4}$$
|
For $f(x) = x^{2} \, e^{-x}$
\begin{align}
f(x) = x^{2} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{n}}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{n+2}}{n!}.
\end{align}
Differentiation leads to
\begin{align}
f'(x) = (2 - x) \, x \, e^{-x} = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, (n+2) \, x^{n+1}}{n!}
\end{align}
Now let $x = 2$ to obtain
\begin{align}
\sum_{n=0}^{\infty} \frac{(-2)^{n} \, (n+2)}{n!} = 0
\end{align}
or, by shifting the first term to the right hand side,
\begin{align}
\sum_{n=1}^{\infty} \frac{(-2)^{n} \, (n+2)}{n!} = -2.
\end{align}
Multiply both sides by $-2$ to obtain
\begin{align}
\sum_{n=1}^{\infty} \frac{(-2)^{n+1} \, (n+2)}{n!} = 4.
\end{align}
|
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|
What is $\lim_{x\to 0} \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$?
What is $\displaystyle\lim_{x\to 0} \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ ?
Find an asymptotic expansion of $\displaystyle \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ as $x\to 0$
One the one hand, $\sqrt{x}\to 0$, but $\displaystyle\frac{\ln n}{1+n^2 x}\sim \ln n$ which suggests that $\sum_{n=2}^\infty \frac{\ln n}{1+n^2 x}$ "diverges" to $\infty$.
I haven't been able to tell which term dominates here, let alone the asymptotic expansion.
Here's the graph of $x\to \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ for $x\in (0,0.5)$
This suggests the limit is $\infty$ and the estimate should be $1/x^\alpha$ for some positive $\alpha$.
|
I know we already have an answer, but I would argue this way:
for all $x$ and $n\ge 2$ we have
$$
\frac{\sqrt{x}\ln{n}}{1+n^2x} > \frac{\sqrt{x}\ln{n}}{n^2(1+x)}
$$
Then
$$
\sum_{n \ge 2} \frac{\sqrt{x}\ln{n}}{1+n^2x} > \frac{\sqrt{x}}{1+x} \sum_{n \ge 2} \frac{\ln{n}}{n^2} > \frac{\sqrt{x}}{1+x} \sum_{n \ge 2} \frac{1}{n^2} = \frac{\sqrt{x}}{1+x}c
$$
Once the series $\sum_{n \ge 2} \frac{1}{n^2}$ is convergent.
This shows you that goes to infinity faster than $\frac{1}{\sqrt{x}}$, when $x \rightarrow 0 $.
|
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|
Why does $\int_0^{2\pi} (1+2\cos(x))/(5+4\cos(x))\,dx$ vanish? The standard substitution $y=\tan(x/2)$ shows that
$$ \int_0^{2\pi} \frac{1+2\cos(x)}{5+4\cos(x)}\,dx = 0. $$
What is the "real explanation" for this fact? My guess is that the "book proof" involves contour integration; is this correct? Is there an elegant "calculus proof" avoiding technical computations?
Thanks!
|
Here is the desired elementary calculus argument. Note first that
$$
\int_0^{2\pi} \frac{1+2\cos x}{5+4\cos x}\,dx \;=\; 2\int_0^\pi \frac{1+2\cos x}{5+4\cos x}\,dx.
$$
Let $u = \arccos\left(-\dfrac{4+5\cos x}{5+4\cos x}\right)$. Note that $u$ decreases continuously from $\pi$ to $0$ as $x$ goes from $0$ to $\pi$. It is easy to check that
$$
1 + 2\cos x\;=\; -3\,\frac{1+2\cos u}{5+4\cos u}\qquad\text{and}\qquad du = \dfrac{-3\,dx}{5+4\cos x},
$$
so
$$
\int_0^\pi \frac{1+2\cos x}{5+4\cos x}\,dx \;=\; \int_\pi^0 \frac{1+2\cos u}{5+4\cos u}\,du \;=\; -\int_0^\pi \frac{1+2\cos u}{5+4\cos u}\,du,
$$
and thus $\displaystyle\int_0^\pi \frac{1+2\cos x}{5+4\cos x}\,dx=0$.
Edit: By the way, the "best" way to evaluate this integral is indeed contour integration. In particular,
$$
\int_0^{2\pi} \frac{1+2\cos x}{5+4\cos x}\,dx \;=\; \oint_C \frac{1+z+z^{-1}}{5+2z+2z^{-1}}\,\frac{dz}{iz} \;=\; \oint_C \frac{z^2+z+1}{iz(z+2)(2z+1)}\,dz,
$$
where $C$ is the unit circle, and the result follows immediately from the residue theorem (since the residue is $-i/2$ at $0$ and $i/2$ at $-1/2$).
Edit 2: The same argument shows in general that
$$
\int_0^\pi \frac{b+(a+c)\cos x}{c + b \cos x}\,dx \;=\; 0.
$$
whenever $a^2+b^2 = c^2$ and $a$, $b$, and $c$ are all positive. The given integral is the case where $a=3$, $b=4$, and $c=5$.
|
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|
Finding the limit of $(1-\cos x)/x^2$ $$\lim _{x \to 0}{1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2$$
now $$\lim_{x \to 0}{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}=\left({\sin\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)}\right)^2=1^2$$
So we have $$\frac{2}{x^2}\cdot \left(\frac{x}{2}\right)^2=\frac{2}{x^2}\cdot \left(\frac{x^2}{4}\right)=\frac{1}{2}$$
Are the moves right?
|
To provide a correction to your own work I would remove the $\lim$ at first because I want to simplifies to the maximum the expression and at the last the computation, as follows:
$${1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2 ={\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}\cdot \frac{1}{2}$$ therefore
$$\lim{1-\cos x\over x^2}=\lim{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}\cdot \frac{1}{2}=1\cdot\frac{1}{2}=\frac{1}{2}.$$
|
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|
Diagonals of parallelepiped have length $2,3,5,11$ Is it possible that the four body diagonals (not face diagonals) of a parallelepiped have lengths $2,3,5$, and $11$?
I guess the answer is no, because it is hard for one diagonal to be longer than even the sum of the remaining diagonals ($11>2+3+5$). Maybe there is a corresponding version of the triangle inequality for parallelepiped?
|
Yes, there is such an inequality. Place the parallelogram with one corner at the origin and the neighboring three corners at $a$, $b$, and $c$. The remaining four corners are at $a+b$, $a+c$, $b+c$ and $a+b+c$.
The length of the body diagonals are
$$ \begin{array}{c} D_0 = |a+b+c| \\ D_1 = |a+b-c| \\ D_2 = |a+c-b| \\ D_3 = |b+c-a| \end{array} $$
Using the triangle equality we get
$$ \begin{align} D_1+D_2+D_3 &= |a+b-c|+|a+c-b|+|b+c-a| \\
& \ge |a+b-c+a+c-b+b+c-a| = |a+b+c| = D_0 \end{align} $$
|
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|
How to calculate the limit: $\lim _{ x\to1 } \frac { nx^{ n+1 }-(n+1)x^{ n }+1 }{ (e^{ x }-e)\sin(\pi x) } $ How to calculate the limit?
$$\lim_{ x\to 1 } \frac { nx^{ n+1 }-(n+1)x^{ n }+1 }{ (e^{ x }-e)\sin(\pi x) } $$
Is binomial theorem needed ?
|
Suppose $n$ is a positive integer first. In that case we can put $x = 1 + h$ and let $h \to 0$. The limit evaluation is as follows
\begin{align}
L &= \lim_{h \to 0}\frac{n(1 + h)^{n + 1} - (n + 1)(1 + h)^{n} + 1}{(e^{1 + h} - e)\sin\pi(1 + h)}\notag\\
&= -\frac{1}{e}\lim_{h \to 0}\frac{n(1 + h)^{n + 1} - (n + 1)(1 + h)^{n} + 1}{(e^{h} - 1)\sin\pi h}\notag\\
&= -\frac{1}{e}\lim_{h \to 0}\frac{n(1 + h)^{n + 1} - (n + 1)(1 + h)^{n} + 1}{h^{2}}\cdot\frac{h}{e^{h} - 1}\cdot\frac{\pi h}{\sin\pi h}\cdot\frac{1}{\pi}\notag\\
&= -\frac{1}{\pi e}\lim_{h \to 0}\frac{n(1 + h)^{n + 1} - (n + 1)(1 + h)^{n} + 1}{h^{2}}\tag{1}\\
&= -\frac{1}{\pi e}\lim_{h \to 0}\dfrac{n\left(1 + (n + 1)h + \dfrac{n(n + 1)}{2}h^{2} + \cdots\right) - (n + 1)\left(1 + nh + \dfrac{n(n - 1)}{2}h^{2} + \cdots\right) + 1}{h^{2}}\notag\\
&= -\frac{1}{\pi e}\lim_{h \to 0}\dfrac{\dfrac{n^{2}(n + 1)}{2}h^{2} - \dfrac{n(n^{2} - 1)}{2}h^{2} + o(h^{2})}{h^{2}}\notag\\
&= -\frac{1}{\pi e}\cdot\frac{n^{2} + n}{2} = -\frac{n(n + 1)}{2\pi e}\notag
\end{align} Note that the ellipsis ($\ldots$) used in above derivation represents a finite (because $n$ is positive integer) number of terms in powers of $h$ starting with $h^{3}$ and hence they are combined and written as $o(h^{2})$.
If $n$ is not a positive integer then we have to start from equation $(1)$ and apply L'Hospital's Rule. Thus
\begin{align}
L &= -\frac{1}{\pi e}\lim_{h \to 0}\frac{n(1 + h)^{n + 1} - (n + 1)(1 + h)^{n} + 1}{h^{2}}\notag\\
&= -\frac{1}{\pi e}\lim_{h \to 0}\frac{n(n + 1)(1 + h)^{n} - n(n + 1)(1 + h)^{n - 1}}{2h}\notag\\
&= -\frac{n(n + 1)}{2\pi e}\lim_{h \to 0}\frac{h(1 + h)^{n - 1}}{h}\notag\\
&= -\frac{n(n + 1)}{2\pi e}\notag
\end{align}
|
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|
what is the value of $\int \sin(x)\cos(x)dx$? $\frac{\sin^2(x)}{2}$ or $\frac{-\cos^2(x)}{2}$ or $\frac{-\cos(2x)}{4}$ $\int \sin(x)\cos(x)dx = \frac{\sin^2(x)}{2}$ because
$$\frac{d}{dx}\frac{\sin^2(x)}{2}=\sin(x)\frac{\sin(x)}{dx}=\sin(x)\cos(x)$$
but also
$$\frac{d}{dx}\frac{-\cos^2(x)}{2}=-\cos(x)\frac{\cos(x)}{dx}=\sin(x)\cos(x)$$
Moreover
$$\int \sin(x)\cos(x)dx = \int \frac{\sin(2x)}{2}dx = \frac{-\cos(2x)}{4} = \frac14 - \frac{\cos^2(x)}{2} = \frac{\sin^2(x)}{2} - \frac14 $$
It's true that $\frac{\sin^2(x)}{2}$ and $\frac{-\cos^2(x)}{2}$ are not equal so what is that problem I made here.
|
$\displaystyle \int \sin x\cos xdx = \displaystyle \int \dfrac{\sin 2x}{2}dx = -\dfrac{\cos 2x}{4}+C$
|
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|
How can I find the remainder? How can I find the remainder when
$$(12371^{56}+34)^{28}$$
is divided by $111$.
I have tried congruences modulo $111$ but without any success.
|
$111=3\cdot 37$. You can use Euler's criterion (EC) with Quadratic reciprocity.
$$\left(\frac{13}{37}\right)=\left(\frac{37}{13}\right)=\left(\frac{-2}{13}\right)=-1$$
$$12371^{56}\equiv 13^{56}\equiv 13^2\left(13^{(37-1)/2}\right)^3\stackrel{\text{EC}}\equiv 21\left(-1\right)^3\equiv 16\pmod{\! 37}$$
$$\left(12371^{56}+34\right)^{28}\equiv 13^{28}\stackrel{\text{EC}}\equiv -13^{10}\equiv -\left(13^5\right)^2\equiv -4\pmod{\! 37}$$
because $13^5\equiv 21^2\cdot 13\equiv -3\cdot 13\equiv -2\pmod{\! 37}$.
$$\left(12371^{56}+34\right)^{28}\equiv \left((-1)^{56}+1\right)^{28}\equiv (-1)^{28}\equiv 1\pmod{\! 3}$$
$x\equiv -4\equiv 70\pmod{\! 37},\, x\equiv 1\equiv 70\pmod{\! 3}$ and
$$3,37\mid x-70\iff 111\mid x-70$$
|
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|
Prove that $\tan^6 20°+\tan^6 40°+\tan^6 80°$ is an integer Prove that $\tan^6 20°+\tan^6 40°+\tan^6 80°$ is an integer. Doesn't this problem seem a little out of the box? It seems beautiful, but I don't have an idea on how to start. Calculating the value does give me an integer-$32733$. Thanks.
|
Let $\theta = 20^\circ = \frac{\pi}{9}$ and let $\omega_k = \tan(k\theta)$ for $k = 0,1,\ldots 8$. The sum we want can be rewritten as
$$\tan^6(20^\circ) + \tan^6(40^\circ) + \tan^6(80^\circ) = \omega_1^6 + \omega_2^6 + \omega_4^6$$
Since
$(\cos(k\theta) + i\sin(k \theta))^9 = e^{i9k\theta} = (-1)^k$,
the $\omega_k$ are the nine roots of the polynomial
$$
\Im\left[( 1 + i t )^9\right]
= t^9-36t^7+126t^5-84t^3+9t
= t(t^2-3)(t^6-33t^4+27t^2-3)
$$
Notice $\omega_0 = 0, \omega_3 = -\omega_6 = \sqrt{3}$, the factor
$t(t^2 - 3) = (t - \omega_0)(t - \omega_3)(t - \omega_6)$.
Together with the identities $\omega_1 = -\omega_8$, $\omega_2 = -\omega_7$, $\omega_4 = -\omega_5$, we find
$$(x - \omega_1^2)(x - \omega_2^2)(x - \omega_4^2) = x^3 - 33x^2 + 27 x - 3$$
Let $p_n = \omega_1^{2n} + \omega_2^{2n} + \omega_3^{2n}$ for $n \in \mathbb{Z}_{+}$. Apply Newton's identities to above cubic polynomial, we have
$$
\begin{cases}
p_1 - 33 = 0\\
p_2 - 33 p_1 + 2\times 27 = 0\\
p_3 - 33 p_2 + 27 p_1 - 3\times 3 = 0
\end{cases}
\implies
\begin{cases}
p_1 = 33\\
p_2 = 33 \times 33 - 2\times 27 = 1035\\
p_3 = 33 \times 1035 - 27 \times 33 + 3\times 3 = 33273
\end{cases}
$$
As a result,
$$\tan^6(20^\circ) + \tan^6(40^\circ) + \tan^6(80^\circ) = \omega_1^6 + \omega_2^6 + \omega_4^6 = p_3 = 33273$$
|
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|
Compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ where $\omega^3 = 1$ If $\omega^3 = 1$ and $\omega \neq 1$, then compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$
I'm pretty lost, I don't really know where to start.
Thanks
|
Let $P$ be the product, then: $P = 1^2-(\omega - \omega^2)^2=1-(\omega^2-2\omega^3+\omega^4)=1-\omega^2+2-\omega^4=3-\omega^2-\omega$. Observe that $1-\omega^3= (1-\omega)(1+\omega+\omega^2) = 0$ since $1 \neq \omega$. So: $-\omega - \omega^2 = 1$, and $P = 3-(-1) = 4$.
|
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|
Calculus 2 - $\int(\sqrt{72+36x^2}dx$ I have done this problem several times and this is the only answer i ever come to. My schools webwork gives me incorrect for my answer (answer is not simplified but it should be accepted in this format). Did i do this correctly?
Here is my work:
\begin{align}
\int \sqrt{72+36x^2}\, dx&=\sqrt {36}\int \sqrt{2+x^2}\,dx\\
&=6\int \sqrt{2+x^2}\,dx\\
&=6\int \sqrt 2 \sec \theta \sqrt 2 \sec^2 \theta \, d\theta\\
&=12 \int sec^3 \theta=12\left[\frac{\tan \theta \sec \theta}2 +\frac 12 \int \sec\theta \, d\theta\right]\\
&=12\left[\frac{\tan\theta\sec\theta}2+\frac 12 \ln|\sec\theta+\tan\theta|\right]+C\\
&=6\tan \theta\sec\theta+6\ln|\sec\theta+\tan\theta|+C\\
&=6\tan\left(\tan^{-1}\frac{x}{\sqrt 2}\right)\sec\left(\tan^{-1}\frac{x}{\sqrt 2}\right)\\
&+6\ln \left|\sec\left(\tan^{-1}\frac{x}2\right)+\tan\left(\tan^{-1}\frac{x}{\sqrt 2}\right)\right|+C
\end{align}
Any help is appreciated. Thanks
|
By the above diagram, we can show that
$$
\begin{aligned}
I &=6(\tan \theta \sec \theta+\ln |\sec \theta+\tan \theta|)+c \\
&=6\left(\frac{x}{\sqrt{2}} \cdot \frac{\sqrt{x^{2}+2}}{\sqrt{2}}+\ln \left|\frac{x}{\sqrt{2}}+\frac{\sqrt{x^{2}+2}}{\sqrt{2}}\right|\right)+c \\
&=3x \sqrt{x^{2}+2}+6 \ln \left|x+\sqrt{x^{2}+2}\right|+C
\end{aligned}
$$
:|D Wish it can help!
|
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|
Complex numbers - roots of unity
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}.$$
I have tried adding the first two and the second two separately, then adding those sums but how do I get a numerical value as the answer?
Thanks
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Maybe more direct
$$
\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}
=
\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^{-1}} + \frac{\omega^{3}}{1 - \omega} + \frac{\omega^{-1}}{1 - \omega^{-2}}
$$
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|
System of equations in a,b,c,d $a,b,c,d$ are complex numbers satisfying
\begin{cases}
a+b+c+d=3 \\
a^2+ b^2+ c^2+ d^2=5 \\
a^3+ b^3+ c^3+ d^3=3 \\
a^4+ b^4+ c^4+ d^4=9
\end{cases}
Find the value of the following: $$a^{2015} + b^{2015} + c^{2015} + d^{2015}$$.
|
$a^2+b^2=(a+b)^2-2ab=(3-(c+d))^2-2ab \implies ab=2-3(c+d)+(c+d)^2-cd$
$a^3+b^3=(a+b)^3-3ab(a+b)=(3-(c+d))^3-3(3-(c+d))(2-3(c+d)+(c+d)^2-cd)$
$c^3+d^3=(c+d)^3-3cd(c+d)$
$u=c+d,v=cd \implies (3-u)^3-3(3-u)(2-3u+u^2-v)+u^3-3uv=3 \implies v=\dfrac{u^3-3u^2+2u+2}{2u-3} $
$a^4+b^4+c^4+d^4=(a^2+b^2)^2-2(ab)^2+((c+d)^2-2cd)^2-2(cd)^2=(5-u^2+2v)^2-2(u^2-3u+2-v)^2+(u^2-2v)^2-2v^2=9$
$v^2-u^2v-3uv+7v+3u^3-9u^2+6u+2=0$ ,replace $v$ we get:
$(u-2)(u-1)(u^2+1)(u^2-6u+10)=0$
$u_1=1,u_2=2,u_3=i,u_4=3-i,u_5=-i,u_6=3+i$
note: $u_1+u_2=u_3+u_4=u_5+u_6=3$ which means the root is simply different combination of $a,b,c,d$, so we only take one root is enough for final answer,take $u=1 \implies v=-2\implies c=-1,d=2 ,a=1+i,b=1-i$
so the final answer is easy to calculate.
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|
$5^m$, where m is any natural, can be expressed as the sum of two perfect squares? Prove that for all natural $m$, $5^m$ can be expressed as the sum of two perfect squares.
Also, prove that $5^m + 2$ can be expressed as the sum of three perfect squares.
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You can prove the first statement by induction on $m$. $m = 1$: $5^1 = 5 = 1+4 = 1^2+2^2$. Assume $5^m = a^2+b^2$, we have: $5^{m+1} = (1^2+2^2)(a^2+b^2) = (a-2b)^2+(b+2a)^2$ is a quite well-known identity.
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|
Why is $\max(x, x')$ equivalent to $\frac{1}{2}( x + x' + |x - x' |)$? Why is it that
$$\max(x, x') = \frac{1}{2}( x + x' + |x - x'|)$$
is true? Is it supposed to be obvious? Because it seems to come out of thin air for me. Anyway, I've verified this by plotting it in matlab, so it works, but its hardly a satisfying proof/justification.
I've actually have been shown a proof but I am unable to internalize and understand it properly. I will provide it below:
$$\begin{align*}
\max(x, x') &= \max\left(x - \frac{1}{2}(x + x'),\; x' - \frac{1}{2}(x + x')\right) + \frac{1}{2}(x + x')\\\\
&=\max\left(\frac{1}{2}(x - x'), \frac{1}{2}(x' - x)\right) + \frac{1}{2}(x + x')\\\\
&=\max\left(\frac{1}{2}(x' - x), \frac{1}{2}(x - x')\right) + \frac{1}{2}(x + x')\\\\
&=\max\left(-\frac{1}{2}(x - x'), \frac{1}{2}(x - x')\right) + \frac{1}{2}(x + x')\\\\
&=\left| \frac{1}{2}(x - x')\right| + \frac{1}{2}(x + x') \\\\
&= \frac{1}{2}( x + x' + |x -x'| )
\end{align*}$$
The only line that remains a mystery for me is actually the first line.
$$ \max(x, x') = \max\left(x - \frac{1}{2}(x + x'),\; x' - \frac{1}{2}(x + x')\right) + \frac{1}{2}(x + x') $$
After that everything seems to be simple algebra manipulations, except for the cool use of the identity
$$\left| \frac{1}{2}(x - x')\right| = \max\left(-\frac{1}{2}(x - x'), \frac{1}{2}(x - x')\right)$$ which makes sense to me.
If anyone has a good explanation of why that step is true, I'd be very thankful. Also, and maybe more importantly, that first step seems like magic to me, I have no idea where it came from and if someone has a good explanation for that too, that would be awesome. I'd like to understand this not only to see why its correct, but also to understand how one would have come up with such a result.
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For determining the maximum of two arguments $x$, $x'$ we need the case distinction which of the two arguments is larger.
For determining the absolute value of one argument $x$ one again needs a case distinction, here if the argument is negative or not.
So it is not that surprising that one can express the maximum in terms of the absolute value.
$$
\lvert x - x'\rvert =
\begin{cases}
x - x' & \mbox{for} & x - x' \ge 0 \iff x \ge x' \\
-(x - x') & \mbox{for} & x - x' < 0 \iff x < x'
\end{cases}
$$
so
\begin{align}
x + x ' + \lvert x - x'\rvert
&=
\begin{cases}
x + x' + x - x' & \mbox{for} & x \ge x' \\
x + x' - (x - x') & \mbox{for} & x < x'
\end{cases}
\\
&=
\begin{cases}
2x & \mbox{for} & x \ge x' \\
2x' & \mbox{for} & x < x'
\end{cases}
\end{align}
which gives
$$
\max(x,x') = \frac{x + x ' + \lvert x - x'\rvert}{2}
$$
Similar one shows
$$
\min(x,x') = \frac{x + x ' - \lvert x - x'\rvert}{2}
$$
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Show a function defined by summation is increasing, another is decreasing Problem: For real numbers $x\ge1$ and $k>0$, let $f:R\rightarrow R$ and $g:R\rightarrow R$ be defined as follows.
$f(x) = -\frac{1}{x}+\sum_{n=1}^{\infty}\frac{1}{(nk+x)^2}$ ,
$g(x)=\frac{1}{2x^2}-\frac{1}{x}+\sum_{n=1}^{\infty}\frac{1}{(nk+x)^2}$.
Show that $f$ is increasing and $g$ is decreasing (that is, $f'(x)\ge0$ and $g'(x)\le0$ ).
I was given this problem to solve. I tried differentiating each function with respect to $x$ and got the following:
$f'(x) = \frac{1}{x^2}-2\sum_{n=1}^{\infty}\frac{1}{(nk+x)^3}$ ,
$g'(x)=-\frac{1}{x^3}+\frac{1}{x^2}-2\sum_{n=1}^{\infty}\frac{1}{(nk+x)^3}$.
I was unable to proceed after this stage. Also i tried the following.
For $x>y\ge1$, i wanted to show that $f(x)-f(y)\ge0$, however, i got
$f(x)-f(y)=\frac{x-y}{xy}+\sum_{n=1}^{\infty}\left[\frac{1}{(nk+x)^2}-\frac{1}{(nk+y)^2}\right]$.
I got a similar expression for $g$. Once again, I am unable to proceed from here.
Your solution(s) to this problem will greatly help me. Thank you in advance.
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We can investigate the monotonicity of the functions $f$ and $g$ as you did because of the following
Lemma. [Fich, 435, Th.7 at p. 438] Let $\{u_n(x)\}$ be a sequence of functions on a segment $[a,b]$ such that for each $n$ a function $u_n’(x)$ is continuous. If a series $\sum_{n=1}^\infty u_n(x)$ converges to $u(x)$ on $[a,b]$ and a series $\sum_{n=1}^\infty f’_n(x)$ uniformly converges to $v(x)$ on $[a,b]$ then $u’(x)=v(x)$.
Let’s establish bounds for the function $$S(k,x)= \sum_{n=1}^{\infty}\frac{1}{(nk+x)^3}=\frac 1{k^3} \sum_{n=1}^{\infty}\frac{1}{(n+x/k)^3}.$$
An upper bound
$$S(k,x)=\frac 1{k^3} \sum_{n=1}^{\infty}\frac{1}{(n+x/k)^3}<$$
$$\mbox{(by area comparison)}$$
$$\frac 1{k^3} \int_{0}^{\infty}\frac{dt}{(t+x/k)^3}=
\frac 1{k^3} \frac{-1}{2(t+x/k)^2}\Big|_0^{\infty}=
\frac 1{k^3}\frac{1}{2(x/k)^2}=\frac{1}{2kx^2}.$$
A lower bound
$$S(k,x)=\frac 1{k^3} \sum_{n=1}^{\infty}\frac{1}{(n+x/k)^3}>$$
$$\mbox{(by area comparison)}$$
$$\frac 1{k^3} \int_{1}^{\infty}\frac{dt}{(t+x/k)^3}=
\frac 1{k^3} \frac{-1}{2(t+x/k)^2}\Big|_1^{\infty}=
\frac 1{k^3}\frac{1}{2(1+x/k)^2}=\frac{1}{2k(k+x)^2}.$$
Therefore since $f'(x) = \frac{1}{x^2}-2S(k,x)$ the function $f$ decreases for $k<1$ and $x>\frac{k^2+k\sqrt{k}}{1-k}$. Nevertheless, for $k\ge 1$ the function $f$ increases, because then $S(k,x)< \frac{1}{2x^2}$.
Since $$g'(x)= -\frac{1}{x^3}+\frac{1}{x^2}-2S(k,x)>-\frac{1}{x^3}+\frac{1}{x^2}-\frac{1}{kx^2}=\frac{kx-k-x}{kx^3},$$ the function $g$ increases for $x>1$ and $k>\frac1{x-1}+1$.
Nevertheless, since $$g'(x)= -\frac{1}{x^3}+\frac{1}{x^2}-2S(k,x)<-\frac{1}{x^3}+\frac{1}{x^2}-\frac{1}{k(k+x)^2}=\frac{x^3(k-1)+x^2(2k^2-k)+x(k^3-2k^2)-k^3}{k(k+x)^2x^3},$$ the function $g$ decreases for sufficiently small $k$.
References
[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, v. II, 7-th edition, M.: Nauka, 1970. (in Russian)
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|
Finding the roots of a different Quadratic equation from the roots of a Given Quadratic equation The Question:
If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$...
Then find the roots of the equation $ax^2-bx(x-1)+c(x-1)^2=0$
My Attempt:
The new equation can be made into a quadratic as:
$$(a-b+c)x^2+(b-2c)x+c=0$$
Now $$\text{Sum of roots}=\dfrac{-b}{a} = \dfrac{2c-b}{a-b+c}$$
And $$\text{Product of roots}=\dfrac{c}{a}=\dfrac{c}{a-b+c}$$
But I don't seem to be going anywhere with the way I'm proceeding
Please Help! Thanks!
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We look at the equation
$$ax^2-bx(x-1)+c(x-1)^2=0.\tag{1}$$
If $a\ne 0$, then $1$ is not a solution, so Equation (1) is equivalent to
$$a\left(\frac{x}{x-1}\right)^2-b\frac{x}{x-1}+c=0.\tag{2}$$
If we put $y=-\frac{x}{x-1}$ then the solutions of (2) are the solutions of $ay^2+by+c=0$. It follows that
$$\frac{-x}{x-1}=\alpha\quad\text{or}\quad \frac{-x}{x-1}=\beta.$$
Now we can easily solve for $x$.
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|
Integrating $\sqrt{1-x^2}$ without using trigonometry I am a beginning calculus student. Tonight I had a thought. Maybe I could calculate $\pi$ using integration, but no trig.
The problem is that I don't really know where to start. I thought perhaps I could use the Chain Rule for derivatives.
$~~~~~~~~~$
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$\pi$ is a transcendental number. You can't get it in a closed form without using transcendental functions. It's not a value of any algebraic expression (polynomial, rational function, root of polynomial,...) evaluated at any algebraic number. So, you either have to bring $\pi$ into it from the boundaries of the integral, or get it from a transcendental function (trigonometric in this case).
Of course, non-closed form is ok, your original integral is a nice example of that. You could develop your function into an infinite series and try to compute integrals term-by-term. That is actually one of the ways to compute $\pi$ numerically: to construct an infinite sum from such integrals.
We can try the expansion in this case. Consider the series
$$(1+z)^{1/2}=1+\frac{1}{2}z-\frac{1}{2!}\frac{1}{2^2}z^2+\frac{1}{3!}\frac{1\cdot 3}{2^3}z^3-\frac{1}{2!}\frac{1\cdot 3\cdot 5}{2^4}z^4+\cdots$$
$$(1-x^2)^{1/2}=1-\frac{1}{2}x^2-\frac{1}{2!}\frac{1}{2^2}x^4-\frac{1}{3!}\frac{1\cdot 3}{2^3}x^6-\frac{1}{4!}\frac{1\cdot 3\cdot 5}{2^4}x^8+\cdots$$
$$\frac{\pi}{4}=\int_0^1 (1-x^2)^{1/2}dx=1-\frac{1}{2}\frac{1}{3}-\frac{1}{2!}\frac{1}{2^2}\frac{1}{5}-\frac{1}{3!}\frac{1\cdot 3}{2^3}\frac{1}{7}-\frac{1}{4!}\frac{1\cdot 3\cdot 5}{2^4}\frac{1}{9}+\cdots$$
$$\frac{\pi}{4}=1-\sum_{n=1}^\infty \frac{(2n-3)!!}{n! 2^n (2n+1)}$$
where $n!!$ is the double factorial (product of every other term) and by definition $(-1)!!=1$.
The convergence is actually not as bad as in the case of Leibniz formula (https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80), you can actually get a reasonable value.
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How to solve this combinations with repetitions problem using generating functions? Find the number of solutions to :
$$x_1 + x_2 + x_3 + x_4 + x_5 = 10$$
where none of the variables can be the number $3$.
I can solve this with Inclusion-Exclusion Principle, but I really love solving this kind of problem with generating functions. I did not manage to solve it with generating functions , this is my try:
I have $5$ variables. None of them can be the number $3$.
Need to find : coefficient of $${x^{10}}$$
so:
*
*$x_1$ can be : $0 , 1 , 2, 4 ,5 , 6, \ldots$ to infinity, that means:
$$1 + x + {x^2} + {x^4} + {x^5} + {x^6} + .....$$
and this is relevant for all of the five variables so thats means total:
$${(1 + x + {x^2} + {x^4} + ...)^5}$$
but I can't find the generating function of this series.
I tried to multiply the series by $x$ and then subtract the original series from the multiplied one
$$\begin{array}{l}1 + x + {x^2} + {x^{{4^{}}}} + ...\\ - {\rm{ }}x + {x^2} + {x^3} + {x^4} + ...\end{array}
$$
and I get:
$${\left( {\frac{{1 - {x^3}}}{{1 - x}}} \right)^5}
$$
but the final solution after I'm using binomial expansion is $1$, and that's not correct.
Can I get help please?
Thanks.
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You can express your close-to-geometric sum as the difference between sums, i.e.
$$f(x)=\sum_{k\ge 0} x^k -x^3=\frac{1}{1-x}-x^3=\frac{1-x^3+x^4}{1-x}$$
Then we have
$$F(x)=\left(\frac{1-x^3+x^4}{1-x}\right)^5=\sum_{j\ge0}\binom{j+4}{4}x^j\cdot\sum_{a+b+c=5}\binom{5}{a,b,c}(-1)^bx^{3b+4c}$$
and equating coefficients we have that $10=j+3b+4c\ \to j=10-3b-4c$ and $j,b,c\ge 0$ and $b+c\le 5\ \to c\le 5-b$. So
$$[x^{10}]F(x)=\sum_{b+c\le 5}\binom{14-3b-4c}{4}\binom{5}{5-b-c,b,c}(-1)^b$$
You can simplify this sum seeing that $14-3b-4c\ge 4$ and so on.
Alternatively you can take the $[x^{10}]$ coefficient from the Maclaurin series of $F(x)$ using some CAS.
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|
How do I prove that $\lim_{(x,y)→(0,0)}\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} = 0$ $$\lim_{(x,y)→(0,0)}\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} = 0$$
I believe this is correct since I couldn't find a directional limit that won't validate this.
From what I know, I have to prove that
$$\forall\epsilon\gt 0, \exists\delta\gt 0$$
$$ \mbox{such that}$$
$$0 \lt \|(x,y)\| \lt \delta\Rightarrow \frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} \lt \epsilon$$
I know that $$\sqrt{x^2+y^2} = \|(x,y)\|$$
So I take $$\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} \le \frac{2}{\|v\|}$$ and I kind of get stuck there. I greatly appreciate some help.
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hint: $$\cos(x^2+y^2) = 1 -\dfrac{(x^2+y^2)^2}{2!}+\dfrac{(x^2+y^2)^4}{4!}-\cdots$$ we are led to the answer as well.
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|
Calculate Infinite Limit I'm trying to calculate the limit and when I get to the last step I plug in infinity for $\frac 8x$ and that divided by -4 I get - infinity for my answer but the book says 0. Where did I go wrong?
$$
\frac {8x^3-x^2}{7+11x-4x^4}
$$
Divide everything by $x^4$
$$
\frac {\frac{8x^3}{x^4}-\frac{x^2}{x^4}}{\frac{7}{x^4}+\frac{11x}{x^4}-\frac{4x^4}{x^4}}
$$
Results
$$
\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4} = \infty
$$
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Continuing from where you've reached, you can conclude that $$\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4}$$
where as $x \to \infty$, we have $\frac{8}{x} \to 0$. The same goes for $\frac{1}{x^2} \to 0$, and $\frac{7}{x^4} \to 0$. We also have $\frac{11x}{x^4} \to 0$, so you can rewrite the above approximately for large $x$ as $$\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4} \approx \frac{0-0}{0+0-4} = 0$$
Alternatively, (this isn't a technique you are likely to understand just yet and I wouldn't recommend using it for the moment, but I am including it any way just for an alternative way), we could use L'Hôpital's rule. Since we have $$\lim_{x\to \infty} \frac {8x^3-x^2}{7+11x-4x^4} = \frac{\infty}{\infty}$$ an indeterminate form L'Hôpital once to get $$\lim_{x\to \infty} \frac {8x^3-x^2}{7+11x-4x^4} = \lim_{x\to \infty}\frac{24x^2 - 2x}{11 - 16x^3}$$
which is still an indeterminate form, apply L'Hôpital again to get $$\lim_{x\to \infty}\frac{24x^2 - 2x}{11 - 16x^3} = \lim_{x\to \infty}\frac{48x - 2}{- 48x^2}$$ still indeterminate, so a final application of L'Hôpital gives us: $$\lim_{x\to \infty}\frac{48x - 2}{- 48x^2} = \lim_{x\to \infty} -\frac{2}{x}$$ which tends to $0$ as $x \to \infty$.
|
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|
Finding the minimum value of a function. Find the minimum value of the function:
$$f(x) = \frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2 - 2}{\left(x + \frac{1}{x}\right)^3 +\left(x^3 + \frac{1}{x^3}\right)}$$
for $x>0$.
I know that this function simplifies into something a little 'nicer' than what we have above. However, I have hit a brick wall after that. Please refrain from using derivatives to find the minimum value.
|
The numerator of $f(x)$ is $$
\begin{align}
\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2=&\Bigg(\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\Bigg)\times
\\&\Bigg(\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\Bigg)\,.
\end{align}$$
Hence, $f(x)=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)=3\left(x+\frac{1}{x}\right)$. The rest is easy, and the answer is that the minimum value of $f(x)$ is $6$, which is obtained only at $x=1$.
|
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|
How to show that that the following three consecutive numbers $3^{2^{10}} − 1$, $3^{2^{10}}$,$3^{2^{10}}+ 1$ are the sum of two squares? Show that the following three consecutive numbers:
$$
3^{2^{10}} − 1, 3^{2^{10}} , 3^{2^{10}} + 1
$$
can be represented as sums of two integer squares.
|
It is known that the set of sums of two squares is closed under the multiplication. On the other hand, it is clear that $3^{2^{10}}$ and $3^{2^{10}} + 1$ satisfy the statement.
Let us prove that in general
$3^{2^{n}} – 1$ is a sum of two squares.
By induction, we have $3^{2^{0}} – 1=2=1+1$ so assume that $$3^{2^{n}} – 1=x^2+y^2$$
We have $$3^{2^{n+1}} – 1=(3^{2^{n}} + 1)( 3^{2^{n}} – 1)$$ It follows
$$3^{2^{n+1}} – 1=(3^{2^{n}} + 1)( 3^{2^{n}} – 1)= (3^{2^{n}} + 1)( x^2+y^2)$$
The first factor is obviously a sum of two squares, therefore $$3^{2^{n+1}} – 1=X^2+Y^2$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.